NEET Practice Questions (MCQs) with Answers & Solutions

L - Energy level = n=2

No. of orbitals = n² = (2)² = 4

$$\begin{gathered} \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{mv}} \\ \mathrm{\lambda } = \frac{1}{\mathrm{\nu }} \left(1\right) \\ \because \mathrm{\lambda } = {\mathrm{\lambda }}_1 \\ \mathrm{now} \mathrm{\nu } = \frac{1}{3}\mathrm{rd} \\ {\mathrm{\lambda }}_2 = \frac{1}{{\displaystyle \frac{\mathrm{\nu }}{3}}} \\ {\mathrm{\lambda }}_2 = \frac{{\displaystyle 3}}{{\displaystyle \mathrm{\nu }}} \\ {\mathrm{\lambda }}_2 = 3·{\mathrm{\lambda }}_1 \left[\mathrm{from} \left(1\right)\right] \\ \% \mathrm{increase} = \frac{{\mathrm{\lambda }}_2-{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_1}\times 100 \\ = \frac{3{\mathrm{\lambda }}_1-{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_1}\times 100 \\ = 2\times 100 \\ = 200\% \end{gathered}$$

$\mu = \sqrt{n(n+2)}$$$

Fe²⁺ $\to$ 1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3d⁴

n=4

Fe³⁺ $\to$1S² 2S² 2P⁶ 3S² 3P⁶ 4S⁰ 3d⁵

n=5

Cr³⁺ $\to$1S² 2S² 2P⁶ 3S² 3P⁶ 4S⁰ 3d³

n=3

Mn⁴⁺ $\to$1S² 2S² 2P⁶ 3S² 3P⁶ 4S⁰ 3d³

n=3

Bohr model is only valid for 1e^- system.

1/$\mathrm{\lambda }$ = RZ² [ 1/n₁² - 1/n₂²] = R x 2² [ 1/1² - 1/2²]

1/$\mathrm{\lambda }$ = R x 4[3/4]; 1/$\mathrm{\lambda }$ =3R; [ $\mathrm{\lambda }$ = 1/3R]

No. of protons $\to$13-7 = 6

No. of protons = Atomic no = 6

10

4.

(d) (a) According to de-Broglie's equation

Wavelength ($\mathrm{\lambda }$) = h/mv

where, h = Planck's constant

Thus, statement (a) is correct.

(b) According to Heisenberg uncertainty principle, the uncertainties of position($∆x$) and momentum (p=m$∆$v) are related as 

$$\begin{gathered} ∆x.∆\mathrm{p}\ge \mathrm{h}/4\mathrm{\pi } \mathrm{or} ∆x.\mathrm{m}∆\mathrm{v}\ge \mathrm{h}/4\mathrm{\pi } \\ ∆x.\mathrm{m}∆\mathrm{a}.∆\mathrm{t}\ge \mathrm{h}/4\mathrm{\pi } [∆\mathrm{v}/∆\mathrm{t} =∆\mathrm{a}, \mathrm{a}=\mathrm{acceleration}] \\ \mathrm{or} ∆x.\mathrm{F}.∆\mathrm{t}\ge \mathrm{h}/4\mathrm{\pi } [{.}_{.}. \mathrm{F} = \mathrm{m}.∆\mathrm{a}] \\ \mathrm{or} ∆\mathrm{E}.∆\mathrm{t}\ge \mathrm{h}/4\mathrm{\pi } [... ∆\mathrm{E} = \mathrm{F}.∆x, \mathrm{E}=\mathrm{energy}] \end{gathered}$$

Thus, statement (b) is correct.

(c) The half and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement. Thus statement (c) is correct.

(d) For a single electronic species like H, energy depends on value of n and does not depend on l. Hence energy of 2s-orbital and 2p-orbital is equal in case of hydrogen like species. Therefore, statement (d) is incorrect.