JEE Advanced 2024 — Paper 1 — CHEM-II solution

JEE Advanced2024Paper 1CHEM-II

Among the following the correct statement(s) for electrons in an atom is(are)

A
Uncertainty principle rules out the existence of definite paths for electrons.
B
The energy of an electron in $2 s$ orbital of an atom is lower than the energy of an electron that is infinitely far away from the nucleus.
C
According to Bohr's model, the most negative energy value for an electron is given by $n = 1$ , which corresponds to the most stable orbit.
D
According to Bohr's model, the magnitude of velocity of electrons increases with increase in values of $n$ .

Solution

(A) Heisenberg's uncertainty principle states that position and momentum of an electron cannot be simultaneously known precisely; a definite trajectory (path) requires both. Hence the very idea of a fixed orbit is ruled out. Correct.

(B) A bound electron in any orbital (including $2 s$ ) has negative energy relative to an electron at infinity (which has zero energy). So the $2 s$ electron has lower energy than a free electron at infinity. Correct.

(C) Bohr's energy expression $E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}}$ eV/atom is most negative for $n = 1$ , corresponding to the most stable (ground) state. Correct.

(D) Bohr's velocity expression $v_{n} = v_{0} \frac{Z}{n}$ shows that velocity decreases as $n$ increases. Incorrect.

$∴$ Correct options: (A), (B), (C).

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