### Question 2.66

NCERT Class XI Chemistry

Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

### Solution in Detail

(video solution below this)

$\displaystyle \underline{\underline{\text{(a) P (phosphorus)}}}$

Atomic number of P is 15

Configuration $\displaystyle \text{[Ne]}3s^23p^3$

but due to exchange energy stabilization half-filled orbitals are more stable

Hence $\displaystyle p_x, p_y, p_z$ each contains 1 unpaired electron

Ans: 3

$\displaystyle \underline{\underline{\text{(b) Si (silicon)}}}$

Atomic number of Si is 14

Configuration $\displaystyle \text{[Ne]}3s^23p^2$

but due to exchange energy stabilization half-filled orbitals are more stable

Hence $\displaystyle p_x, p_y$ each contains 1 unpaired electron

Ans: 2

$\displaystyle \underline{\underline{\text{(c) Cr (chromium)}}}$

Atomic number of Cr is 24

Configuration $\displaystyle \text{[Ar]}4s^13d^5$

but due to exchange energy stabilization half-filled orbitals are more stable

Hence each of the five 3d contains 1 unpaired electron, and 4s also has 1 un-paired

Ans: 6

$\displaystyle \underline{\underline{\text{(d) Fe (iron)}}}$

Atomic number of Fe is 26

Configuration $\displaystyle \text{[Ar]}4s^23d^{6}$

but due to exchange energy stabilization half-filled orbitals are more stable

hence four of the d orbitals will contain 1 electron each and the one of the d orbitals will contain a pair

Ans: 4

$\displaystyle \underline{\underline{\text{(e) Kr (krypton)}}}$

Krypton is a noble gas with fully-filled and paired valence orbitals.

Ans: 0

### Video Explanation

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