Chemistry MCQs for NEET — Practice Questions with Answers

(a) n represents the main energy level and l represents the subshell.

lf n = 4 and l = 3, the subshell is 4f.

In f subshell, there are 7 orbitals and each orbital can accommodate a maximum of two electrons, so, maximum number of elecirons in 4f subshell = 7x 2 = 14

Number of atomic orbitals in an orbit = n² = 4² =16

(a)

 E₁ = 25eV, E₂ = 50eV

$$\begin{gathered} {\mathrm{E}}_1 = \frac{\mathrm{hc}}{{\mathrm{\lambda }}_1} \mathrm{and} {\mathrm{E}}_2 =\frac{\mathrm{hc}}{{\mathrm{\lambda }}_2} \\ \mathrm{or} \frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1} \\ \mathrm{or} \frac{25}{50}=\frac{{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1} \\ \mathrm{or} {\mathrm{\lambda }}_1 = 2{\mathrm{\lambda }}_2 \end{gathered}$$

(d) Key Idea Write the electronic configurations of given ions and find the correct answer.

           Ni³⁺ (28) = [Ar]3d⁷

           Mn³⁺ (25) = [Ar]3d⁴

           Fe³⁺ (26) = [Ar]3d⁵

           Co³⁺ (27) = [Ar]3d⁶

Key Idea Number of oxidation states exhibited by d-block elements is the sum of number of electrons (unpaired) in d-orbitals and number of electrons in s-orbital.

(a) 3d³, 4s² $\Rightarrow$ O.S = 3 + 2 = 5

(b) 3d⁵, 4s¹ $\Rightarrow$ O.S = 5 + 1 = 6

(c) 3d⁵, 4s² $\Rightarrow$ O.S = 5 + 2 = 7

(d) 3d², 4s² $\Rightarrow$ O.S = 2 + 2 = 4

Hence, element with 3d⁵, 4s² configuration exhibits largest number of oxidation states.

Total number of subshells = (2l+1)

$\therefore$Maximum number of electrons m the subshell = 2(2l+1)= 4l+2

Key Idea: For an electron, n may be 0, 1, 2 ... and l=0 to n-1 and m=l to +l (including 0) and s = ± -2

Hence, if n = 3

        l = 0 to (3 -1)

        = 0,1, 2

        m = -l to +l

             = -2-1, 0, +1, +2
Therefore, option (c) is not a permissible set of quantum numbers.

Key Idea : According to Heisenberg’s uncertainly principle, it is impossible to determine simultaneously the position and momentum of a moving microscopic particle, i.e.,

Δx X Δp ≥ h/4$\mathrm{\pi }$

Δx X Δp ≥ h/4$\mathrm{\pi }$

Here Δx = Δp and Δp = m.Δv

Δv² = h/m²4$\mathrm{\pi }$

or Δ=1/2m $\sqrt{\frac{\mathrm{h}}{\mathrm{\pi }}}$

Note : The uncertainty principle in terms of energy and time is given as

ΔE.Δt≥h/4$\mathrm{\pi }$

Key Idea: $∆x \mathrm{x} ∆\mathrm{p}\ge \mathrm{h}/4\mathrm{\pi }$

where, $∆$p = uncertainty in momentum and $∆$v = uncertainty in velocity

Here, $∆$p = 1x10^-18 g cm s^-1

         m = 9x10^-28 g

        $∆$v =$∆$p/m = (1x10^-18)/(9x10^-28) = 1.1 x 10⁹ = 1.0 x10⁹ cm s^-1

(c) The value of l varies from 0 to n-l and the value of m varies from -l to +l through zero. The value of 's' $\pm$ 1/2 which signifies the spin of an electron. The correct sets of quantum number are the following:

         n              l            m             s

(ii)     2             1            1             +1/2

(iv)    1             0            0             -1/2

(v)    3              2            2             +1/2