Atomic Structure MCQs for NEET — Chemistry Questions with Answers

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

$\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.

(b) The element with atomic number, Z = 114 is flerovium (Fl). It is a super heavy artificial chemical element. In the periodic table of the elements, it is transactinide element in the p-block. It is a member of 7th period and is the heaviest known member of the carbon family.

Electronic configuration for Z =114 is

        [Rn]₈₆5f¹⁴,6d¹⁰,7s²,7p²