Atomic Structure MCQs for NEET — Chemistry Questions with Answers

The radius of the Bohr orbit is given by the formula r = n^2 * a0, where a0 is the Bohr radius (0.529 Ã…). For n = 2, the radius is 4 times the Bohr radius. The nuclear charge of Be^3+ is 4, which is the same as that of hydrogen. Hence, the radius of the n = 2 orbit of Be^3+ is the same as the first Bohr orbit of hydrogen.

The number of radial nodes in an orbital is given by the principal quantum number (n) minus the azimuthal quantum number (l) minus 1. For 3s orbital, n = 3, l = 0, so radial nodes = 3 - 0 - 1 = 2. For 2p orbital, n = 2, l = 1, so radial nodes = 2 - 1 - 1 = 0.

(a) 

(C) If l=2, m $\ne$ -3

The electron configuration of Ni (atomic number 28) is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^8 4s^2. When forming the Ni^2+ cation, the two 4s electrons are removed, leaving a configuration of 3d^8. The 3d subshell can hold 10 electrons, so there are 2 unpaired electrons in Ni^2+.

The given electronic configuration 1s^2 2s^2 2p^3 has 3 unpaired electrons, one each in the 2p orbitals. This is because the 2p subshell can hold a maximum of 6 electrons, and with 3 electrons present, there are 3 unpaired electrons.

(C)

The correct order of decreasing energy will be electron 3 which has higher energy than 2 and in the order given.

E1 =  3s

E2 =  4s

E3 =  3d

E4=  3p

1.2 x 10^-12= 9 x 10⁹ x 2x79x[1.6x10^-19]² / r₀

r_{0 = 9 x 79 x 256 x 2 x10^-38 x 10⁹ / 1.2 x 10^-12 = 3 x 10^-14m}

(C) $\lambda =\frac{3\times {10}^{8}m.se{c}^{-1}}{5.09\times {10}^{14}m.se{c}^{-1}}$

=0.5892 x 10^-6 = 589.4 x 10^-9m = 589nm

When an electron is brought closer to the nucleus, the potential energy of the electron-nucleus system decreases due to the attractive force between them. This decrease in potential energy results in a lower total energy of the system.