Hydrides

ChemistryHydrogenFor JEE aspirants

Hydrides are the binary compounds that dihydrogen forms with other elements, written in general as $M H_{x}$ . The type of hydride an element gives depends on where it sits in the periodic table: the s-block gives ionic hydrides, the p-block covalent hydrides and the d- and f-blocks metallic hydrides, with the hydride gap in between. This page covers all three types of hydrides, their structures and properties, and the polymeric and complex hydrides. The topic is in the JEE Advanced syllabus; JEE Main and NEET have dropped it.

On this page1Three types2Ionic3Covalent4Metallic5Hydride gap6Polymeric, complex7Flowchart8Examples

Key Formulas - Quick Reference

A hydride has the general formula $M H_{x}$ , where M is the element and $x$ the number of hydrogen atoms.
★ Must learn Ionic (saline) hydrides: alkali metals and alkaline earth metals except beryllium. $LiH$ , $NaH$ , $Ca H_{2}$ . They contain the hydride ion $H^{-}$ .
★ Must learn Molten ionic hydrides conduct, and hydrogen is set free at the anode: $2 H^{-}$ gives $H_{2} + 2 e^{-}$ .
Hydrolysis: $NaH + H_{2} O$ gives $NaOH + H_{2}$ , and $Ca H_{2} + 2 H_{2} O$ gives $Ca (OH)_{2} + 2 H_{2}$ .
★ Must learn Covalent (molecular) hydrides: p-block elements, general formula $X H_{8 - n}$ where $n$ is the number of outer-shell electrons of X. Group 13 is the exception.
★ Must learn Covalent hydrides by electron count: electron deficient ( $B_{2} H_{6}$ ), electron precise ( $C H_{4}$ ), electron rich ( $N H_{3}$ , $H_{2} O$ , $HF$ ).
Acidic character increases across a period: $N H_{3}$ basic, $H_{2} O$ amphoteric, $HF$ acidic.
Metallic (interstitial) hydrides: d- and f-block, non-stoichiometric: $Ti H_{1.5 - 1.8}$ , $Pd H_{0.6 - 0.8}$ , $La H_{2.87}$ .
★ Must learn The hydride gap: the metals of groups 7, 8 and 9 form no hydrides at all.
Complex hydrides: $LiAl H_{4}$ and $NaB H_{4}$ , both very strong reducing agents.

1. What Hydrides Are

Dihydrogen combines with a large number of elements to give binary compounds called hydrides. Their general formula is $M H_{x}$ , where M is the other element and $x$ is the number of hydrogen atoms. Depending on how the hydrogen is bonded, hydrides fall into three broad categories.

Figure 1: The block an element belongs to decides what kind of hydride it forms.

Exam Trick

Do not memorise lists of compounds. Ask one question instead: how different is the electronegativity? A big gap (metal and hydrogen) means electron transfer and an ionic hydride. A small gap (p-block and hydrogen) means sharing and a covalent hydride. A metal lattice with room to spare means an interstitial hydride.

2. Ionic or Saline Hydrides

These are the hydrides of elements much more electropositive than hydrogen: the alkali metals and the alkaline earth metals, except beryllium. The metal hands its electron over to hydrogen completely, so the compound contains the hydride ion, $H^{-}$ . Common examples are $LiH$ , $NaH$ and $Ca H_{2}$ .

2.1 Characteristics

Crystalline solids, white or greyish in colour.
High melting and boiling points, as expected of an ionic lattice.
High density and a high heat of formation.
They conduct electricity when molten, and the hydrogen comes off at the anode. This is the direct proof that hydrogen is present as a negative ion.

Ca H_{2} (melt) C a^{2 +} + 2 H^{-}

2 H^{-} anode H_{2} + 2 e^{-}

C a^{2 +} + 2 e^{-} cathode Ca

Figure 2: Hydrogen comes off at the anode. That can only happen if it travelled as the negative ion

H^{-}

, which is the proof that these hydrides are ionic.

They react vigorously with water and with other protic solvents, releasing dihydrogen. The hydride ion is a very strong base, so it snatches a proton from anything that has one.

NaH + H_{2} O NaOH + H_{2}

LiH + C H_{3} OH LiOC H_{3} + H_{2}

NaH + N H_{3} NaN H_{2} + H_{2}

Ca H_{2} + 2 H_{2} O Ca (OH)_{2} + 2 H_{2}

The products here are lithium methoxide and sodamide. $Ca H_{2}$ reacts so cleanly with water that it has its own name, hydrolith, and is carried as a portable source of dihydrogen.

The hydride ion is a proton acceptor, so every one of these reactions is an acid-base reaction, not a redox reaction in the usual sense. The dihydrogen released contains one hydrogen from the hydride and one from the water or alcohol.

Key idea

Ionic hydrides contain

H^{-}

: they conduct when molten with

H_{2}

at the anode, and they react with water, alcohols and ammonia to give

H_{2}

Quick Recall: tap to check

Which electrode gives

H_{2}

when molten

Ca H_{2}

is electrolysed?

The anode, because hydrogen travels as the negative ion

H^{-}

Why is

Be H_{2}

not an ionic hydride?

Beryllium is small and fairly electronegative, so it cannot give up its electrons fully; its hydride is polymeric,

(Be H_{2})_{n}

What is hydrolith?

Calcium hydride,

Ca H_{2}

, a portable source of

H_{2}

3. Covalent or Molecular Hydrides

These are the hydrides of the p-block elements, whose electronegativity is close to that of hydrogen. Neither atom can take the electron away from the other, so they share: $C H_{4}$ , $N H_{3}$ , $H_{2} O$ , $HF$ , $P H_{3}$ and so on.

The formula follows a simple rule. If X has $n$ electrons in its outer shell, the hydride is $X H_{8 - n}$ :

Group	Outer electrons, $n$	Formula $X H_{8 - n}$	Example
14	$4$	$X H_{4}$	$C H_{4}$ , $Si H_{4}$
15	$5$	$X H_{3}$	$N H_{3}$ , $P H_{3}$
16	$6$	$X H_{2}$	$H_{2} O$ , $H_{2} S$
17	$7$	XH	$HF$ , $HCl$

Group 13 is the exception. Its elements have only three valence electrons, one short of what a normal covalent structure needs, so they form polynuclear electron-deficient hydrides such as $B_{2} H_{6}$ and $G a_{2} H_{6}$ instead of simple $B H_{3}$ .

JEE Advanced

Why diborane needs bridges. $B_{2} H_{6}$ has only $2 \times 3 + 6 \times 1 = 12$ valence electrons, but an ethane-like structure would need $14$ for seven two-electron bonds. The four terminal $B - H$ bonds are ordinary; they use $8$ electrons and lie in one plane. The remaining $4$ electrons form two three-centre two-electron bonds: each pair holds $B - H - B$ together, above and below that plane. These banana bonds are longer ( $133 pm$ against $119 pm$ ) and weaker, which is why diborane reacts so readily, for example with water to give boric acid.

Figure 3: Diborane has only

12

valence electrons. Eight make the four normal

B - H_{t}

bonds; the last four make two banana bonds, each one pair spread over

B - H - B

3.1 Sorting them by electron count

Figure 4: Counting electrons sorts every covalent hydride into one of three families, and the family predicts the bonding.

Family	Group	Example	What is special
Electron deficient	13	$B_{2} H_{6}$	fewer electrons than a normal structure needs, so two hydrogen atoms bridge the two boron atoms
Electron precise	14	$C H_{4}$	exactly the electrons needed, four bonds and no lone pair
Electron rich	15, 16, 17	$N H_{3}$ , $H_{2} O$ , $HF$	spare lone pairs, so these hydrides form hydrogen bonds

The electron rich hydrides stand out as soon as their boiling points are plotted. Their lone pairs and very polar bonds let them form hydrogen bonds, so the period-2 member of groups 15, 16 and 17 boils far above the trend of its heavier neighbours:

Figure 5: Only the electron rich hydrides break the trend:

N H_{3}

H_{2} O

and

HF

boil far above the next member of their group because of hydrogen bonding, while electron precise

C H_{4}

does not.

3.2 Characteristics

They exist as separate molecules held together only by weak van der Waals forces, so they are soft solids or gases with low melting and boiling points.
They are poor conductors of electricity, having no ions.
They dissolve in organic solvents rather than in water, being covalent.
They decompose on heating into their elements.
$Si H_{4} heat Si + 2 H_{2}$
Some of them react with water and liberate dihydrogen.
$B_{2} H_{6} + 6 H_{2} O 2B (OH)_{3} + 6 H_{2}$

The product here is boric acid.
Acidic character increases from left to right along a period, as the electronegativity of the central atom rises.

Hydride $N H_{3}$ $H_{2} O$ $HF$

Nature Basic Amphoteric Acidic

Hydride	$N H_{3}$	$H_{2} O$	$HF$
Nature	Basic	Amphoteric	Acidic

Key idea

Count the valence electrons: group 13 hydrides are electron deficient and bridge, group 14 hydrides are precise, and groups 15 to 17 are electron rich and hydrogen bond.

Ionic hydridess-block metals except Be
contain the

H^{-}

ion
crystalline, high melting
conduct only when molten

Covalent hydridesp-block elements
separate molecules
gases or volatile liquids
do not conduct

4. Metallic or Interstitial Hydrides

These are formed by the d-block and f-block metals. Hydrogen atoms are not bonded in the ordinary sense at all: they simply occupy the interstitial spaces, the empty gaps in the metal lattice. Some chemists describe them as solid solutions of hydrogen in the metal rather than as compounds.

4.1 Which metals form them, and the hydride gap

Transition metals of groups 3, 4, 5 and 6 of the d-block.
All the f-block elements.
Some of groups 10 to 12, notably nickel and palladium.

Figure 6: Groups 7, 8 and 9 form no hydrides at all. This blank region is called the hydride gap.

Hydride gap: the metals of groups 7, 8 and 9 do not form hydrides at all. This blank region in the middle of the d-block is called the hydride gap.

4.2 Non-stoichiometric composition

Because the hydrogen atoms only fill gaps, the number that fit depends on temperature and pressure rather than on valency. The composition is therefore not a simple whole number ratio, and these compounds are called non-stoichiometric hydrides.

Hydride	Composition	Metal
$Ti H_{x}$	$x = 1.5$ to $1.8$	titanium
$Zr H_{x}$	$x = 1.30$ to $1.75$	zirconium
$V H_{0.56}$	fixed at about $0.56$	vanadium
$Pd H_{x}$	$x = 0.6$ to $0.8$	palladium
$La H_{2.87}$	about $2.87$	lanthanum
$Yb H_{2.55}$	about $2.55$	ytterbium

4.3 Characteristics

They look like metals: grey or black powders and brittle solids with a metallic appearance.
They conduct electricity, though the conductivity falls as the temperature rises, exactly as in a metal.
They have high thermal conductivity.
They are harder than the parent metal, because the trapped hydrogen atoms stop the layers of metal from sliding.
They decompose reversibly into the metal and dihydrogen on heating. This last property is the one that matters: it turns the metal into a hydrogen sponge.

Figure 7: Ions in a lattice, separate molecules, or hydrogen atoms parked in the gaps of a metal. The structure explains every property that follows.

Exam Trick

The reversible decomposition is why alloys such as Fe-Ti and $NaN i_{5}$ are proposed for storing hydrogen in vehicles. The metal soaks up hydrogen when it is available and gives it back on gentle heating, which is far safer than a tank of compressed gas.

The catch is the amount of hydrogen held per kilogram. The chart compares hydrides used or proposed for storage:

Figure 8: Interstitial hydrides release their hydrogen easily but hold very little of it by mass (

1.91%

for

FeTi H_{2}

). Light hydrides such as

Mg H_{2}

(

7.66%

) hold far more, which is the trade-off in storage.

Quick Recall: tap to check

Which groups make up the hydride gap?

Groups 7, 8 and 9 of the d-block.

Why are interstitial hydrides non-stoichiometric?

Hydrogen only fills holes in the metal lattice, so the amount depends on temperature and pressure, not on valency:

Ti H_{1.5 - 1.8}

Pd H_{0.6 - 0.8}

Which is more acidic,

N H_{3}

HF

HF

: acidity rises across a period with the electronegativity of the central atom.

5. Two More Families

5.1 Polymeric hydrides

When the electronegativity of the element lies between about $1.4$ and $2.0$ , the bonding is neither properly ionic nor properly molecular. The units link up through hydrogen bridges into chains or sheets, giving polymeric hydrides such as $(Be H_{2})_{n}$ , $(Al H_{3})_{n}$ and $(Si H_{2})_{n}$ . They are amorphous solids, stable up to about $525 K$ , above which they begin to give off hydrogen.

This is why $Be H_{2}$ is not grouped with $Mg H_{2}$ and $Ca H_{2}$ . Beryllium is small and relatively electronegative, so it cannot hand its electrons over completely and its hydride comes out polymeric instead of ionic.

5.2 Complex hydrides

These contain the hydride ion coordinated to a central metal atom inside a complex anion. The two that matter are lithium aluminium hydride, $LiAl H_{4}$ , which contains the $[Al H_{4}]^{-}$ ion, and sodium borohydride, $NaB H_{4}$ , which contains $[B H_{4}]^{-}$ . Both are very powerful reducing agents and are used constantly in organic chemistry to reduce carbonyl compounds to alcohols.

Figure 9: In

[B H_{4}]^{-}

and

[Al H_{4}]^{-}

four hydrides sit tetrahedrally round the central atom. Each can deliver

H^{-}

to a carbonyl carbon, which is why both reduce aldehydes and ketones to alcohols.

NaB H_{4}

milder reducing agent
can be used in water or ethanol
reduces aldehydes and ketones
leaves esters and acids alone

LiAl H_{4}

much stronger reducing agent
reacts violently with water: dry ether
reduces aldehydes and ketones
also acids and esters, to alcohols

5.3 Where hydrides are used

Drying agents. Ionic hydrides remove the last traces of water from organic solvents.
Reducing agents. $LiAl H_{4}$ and $NaB H_{4}$ in the laboratory, $NaH$ for making sodamide.
Portable hydrogen. $Ca H_{2}$ , or hydrolith, releases dihydrogen on contact with water.
Hydrogen storage. Interstitial hydrides of Fe-Ti and $NaN i_{5}$ hold hydrogen safely and release it when heated.

Key idea

Polymeric hydrides link through hydrogen bridges; complex hydrides carry

H^{-}

inside an anion and are the standard reducing agents of organic chemistry.

6. Classifying a Hydride: Flowchart and Mind Map

To predict the hydride of any element, ask where it sits in the periodic table. The flowchart asks the three questions in order; the mind map after it puts the whole page on one screen.

Figure 10: Flowchart: three questions about position in the periodic table decide the hydride type, and the type predicts every property on this page.

Figure 11: Mind map: the whole page on one screen. The electronegativity gap between hydrogen and its partner decides the branch.

7. Solved Examples

Solved Example 1

Ionic hydrides are frequently used to remove traces of water from organic compounds. What is the underlying basis of this process?

Solution:

The hydride ion $H^{-}$ is an extremely strong Bronsted base: it accepts a proton from almost anything. Water is the strongest acid present in a damp organic solvent, so the hydride attacks the water and nothing else.

Ca H_{2} + 2 H_{2} O Ca (OH)_{2} + 2 H_{2}

Answer: the water is destroyed, not merely absorbed. The products are an insoluble hydroxide and dihydrogen gas, both of which leave the solvent easily, so the drying is complete rather than an equilibrium.

Solved Example 2

Complete and balance: (i)

NaH + H_{2} O

(ii)

LiH + C H_{3} OH

(iii)

NaH + N H_{3}

Solution:

In each case the hydride ion takes a proton from the other reactant, so dihydrogen is released and the remaining anion pairs with the metal.

NaH + H_{2} O NaOH + H_{2}

LiH + C H_{3} OH LiOC H_{3} + H_{2}

NaH + N H_{3} NaN H_{2} + H_{2}

Answer: sodium hydroxide, lithium methoxide and sodamide, with dihydrogen in every case. Notice that one hydrogen atom of the $H_{2}$ comes from the hydride and the other from the acid.

Solved Example 3

Why is the region of groups 7, 8 and 9 called the hydride gap?

Solution:

Metals of these three groups, such as manganese, iron and cobalt, do not combine with dihydrogen to give hydrides under normal conditions. Their lattices hold the atoms too tightly for hydrogen to enter the interstitial spaces, and the metals are not electropositive enough to give up electrons to hydrogen.

Answer: because a blank strip runs down the middle of the d-block where no hydrides form at all. Groups 3 to 6 on one side and groups 10 to 12 on the other do form them, so the gap stands out clearly.

Solved Example 4

A titanium hydride has the composition

Ti H_{1.8}

. Calculate the percentage by mass of hydrogen in it. (Ti

= 47.9

, H

= 1.0

)

Solution:

Take one mole of the formula unit.

M = 47.9 + 1.8 \times 1.0 = 49.7 g mol^{- 1}

% H = \frac{1.8}{49.7} \times 100 = 3.62%

Answer: about $3.6%$ . The fraction $1.8$ is not a whole number because the hydrogen only fills gaps in the lattice, so the compound is non-stoichiometric.

Solved Example 5

Arrange

N H_{3}

H_{2} O

and

HF

in increasing order of acidic character and give the reason.

Solution:

All three lie in the same period, and the electronegativity of the central atom rises from nitrogen to oxygen to fluorine. The more electronegative the central atom, the more it pulls the bonding electrons away from hydrogen, and the more easily that hydrogen leaves as $H^{+}$ .

Answer: $N H_{3}$ (basic) less acidic than $H_{2} O$ (amphoteric) less acidic than $HF$ (acidic). The same trend holds for any period of covalent hydrides.

Solved Example 6

Which of the following is an electron deficient hydride?
(A)

C H_{4}

(B)

B_{2} H_{6}

(C)

N H_{3}

(D)

H_{2} O

Solution:

Answer: (B). Boron has only three valence electrons, so $B H_{3}$ would be one pair short of a stable structure. The molecule solves the problem by pairing up: two boron atoms share two bridging hydrogen atoms in $B_{2} H_{6}$ .

$C H_{4}$ is electron precise, and $N H_{3}$ and $H_{2} O$ are electron rich, with one and two lone pairs.

Solved Example 7

Calculate the volume of dihydrogen at STP obtained when

4.2 g

of hydrolith reacts completely with water. (Ca

= 40

, H

= 1

)

Solution:

Hydrolith is $Ca H_{2}$ , molar mass $40 + 2 = 42 g mol^{- 1}$ .

Ca H_{2} + 2 H_{2} O Ca (OH)_{2} + 2 H_{2}

n_{Ca H_{2}} = \frac{4.2}{42} = 0.1 mol \Rightarrow n_{H_{2}} = 0.2 mol

V = 0.2 \times 22.4 = 4.48 L at STP

Answer: $4.48 L$ . Each formula unit gives two molecules of $H_{2}$ , one hydrogen from the hydride and one from the water.

Solved Example 8

Which of the following is an electron rich hydride with the highest boiling point in its group?
(A)

C H_{4}

(B)

H_{2} O

(C)

B_{2} H_{6}

(D)

Si H_{4}

Solution:

Answer: (B). Water has two lone pairs on oxygen, so it is electron rich and forms hydrogen bonds; it boils at $373 K$ against $271 K$ for $H_{2} Te$ , the next highest in group 16. $C H_{4}$ and $Si H_{4}$ are electron precise and $B_{2} H_{6}$ is electron deficient.

Solved Example 9

Magnesium hydride releases hydrogen on heating:

Mg H_{2} \to Mg + H_{2}

. Calculate the percentage of hydrogen by mass in

Mg H_{2}

and the volume of

H_{2}

at STP released by

1.00 kg

of it. (Mg

= 24.3

, H

= 1.0

)

Solution:

Molar mass $= 24.3 + 2.0 = 26.3 g mol^{- 1}$ .

% H = \frac{2.0}{26.3} \times 100 = 7.6%

n = \frac{1000}{26.3} = 38.0 mol \Rightarrow V = 38.0 \times 22.4 = 851 L

Answer: $7.6%$ hydrogen, and about $851 L$ of $H_{2}$ at STP from one kilogram, four times as much as $FeTi H_{2}$ would give.

Practice Questions

When an electric current is passed through an ionic hydride in the molten state,
(A) hydrogen is liberated at the anode
(B) hydrogen is liberated at the cathode
(C) no reaction takes place
(D) the hydride ion migrates towards the cathodeAnswer: (A). The hydride ion is negative, so it travels to the anode and is discharged there as $H_{2}$ .
Which of the following is an ionic hydride?
(A) $As H_{3}$ (B) $Si H_{4}$ (C) $Ca H_{2}$ (D) $H_{2} O$ Answer: (C). Calcium is strongly electropositive, so it transfers electrons and gives $C a^{2 +}$ and $H^{-}$ . The other three are covalent.
Why is $Be H_{2}$ not an ionic hydride even though beryllium belongs to group 2?Answer: Beryllium is small and comparatively electronegative, so it cannot transfer its electrons completely. Its hydride is polymeric, $(Be H_{2})_{n}$ , with hydrogen bridges.
Give one reason why interstitial hydrides are useful for storing hydrogen.Answer: They decompose reversibly: the metal absorbs hydrogen and releases it again on gentle heating, which is much safer than storing the compressed gas.
Name two complex hydrides used as reducing agents and give the anion each contains.Answer: $LiAl H_{4}$ , which contains $[Al H_{4}]^{-}$ , and $NaB H_{4}$ , which contains $[B H_{4}]^{-}$ .
Why do $N H_{3}$ , $H_{2} O$ and $HF$ boil far above the next hydride of their groups?Answer: They are electron rich, with lone pairs on a small, very electronegative atom, so their molecules are held together by hydrogen bonds.
How many valence electrons does $B_{2} H_{6}$ have, and how many hold the two bridges?Answer: $12$ in all; $8$ are in the four terminal B-H bonds and $4$ in the two three-centre bridges.

Common Mistakes to Avoid

Watch out

Assuming every metal forms a hydride. Groups 7, 8 and 9 form none at all, which is the hydride gap.
Sending the hydride ion to the cathode. $H^{-}$ is negative, so it moves to the anode and hydrogen is released there.
Calling $Be H_{2}$ an ionic hydride because beryllium is in group 2. It is polymeric, and beryllium is the standard exception.
Giving interstitial hydrides neat whole-number formulae. Their composition varies with temperature and pressure, as in $Ti H_{1.5 - 1.8}$ .
Treating $LiAl H_{4}$ and $NaB H_{4}$ as simple binary hydrides. They are complex hydrides built around the $[Al H_{4}]^{-}$ and $[B H_{4}]^{-}$ ions.
Assuming every covalent hydride is acidic. Along one period $N H_{3}$ is basic, $H_{2} O$ is amphoteric and only $HF$ is acidic.
Writing $B H_{3}$ as the hydride of boron. Group 13 is electron deficient, so the stable hydride is the dimer $B_{2} H_{6}$ .
Saying ionic hydrides dissolve in water. They react with it, giving the hydroxide and dihydrogen, which is a very different thing.

Frequently Asked Questions

What are hydrides?

Hydrides are binary compounds of hydrogen with another element, written in general as $M H_{x}$ . They fall into three main classes: ionic hydrides of the s-block, covalent hydrides of the p-block and metallic or interstitial hydrides of the d- and f-blocks.

What is the hydride gap?

The metals of groups 7, 8 and 9 of the d-block, such as manganese, iron and cobalt, form no hydrides at all. This blank strip running down the middle of the periodic table is called the hydride gap. The groups on either side do form interstitial hydrides.

Why are ionic hydrides used to dry organic solvents?

The hydride ion is a very strong base, so it takes a proton from any water present and destroys it, giving the metal hydroxide and dihydrogen. The water is removed completely rather than merely absorbed, and both products leave the solvent easily.

How do we know hydrogen is present as the hydride ion in sodium hydride?

By electrolysing the molten hydride. Hydrogen is liberated at the anode, the positive electrode. Only a negatively charged species travels to the anode, so the hydrogen must have been present as $H^{-}$ and not as a proton.

What are non-stoichiometric hydrides?

Interstitial hydrides in which the hydrogen simply occupies gaps in the metal lattice, so the number of hydrogen atoms is not a whole-number ratio and varies with temperature and pressure. Examples are $Ti H_{1.5 - 1.8}$ , $Pd H_{0.6 - 0.8}$ and $La H_{2.87}$ .

Why is diborane called an electron deficient hydride?

Boron has three valence electrons, one fewer than a normal covalent structure needs. $B_{2} H_{6}$ therefore has too few electrons for ordinary two-centre bonds, and it solves the problem by letting two hydrogen atoms bridge the two boron atoms in three-centre bonds.

Are hydrides asked in NEET?

Not as a topic of their own, because NMC removed the Hydrogen chapter from the NEET syllabus. The ideas still appear in other chapters NEET does test: hydrogen bonding in water and ammonia, diborane in the p-block and hydride reducing agents in organic chemistry.

Are hydrides in the JEE syllabus?

In JEE Advanced, yes: the syllabus lists ionic, covalent and interstitial hydrides under the Hydrogen unit. JEE Main no longer has the Hydrogen chapter, though diborane and the reducing agents $LiAl H_{4}$ and $NaB H_{4}$ still come up through the p-block and organic chemistry.

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