Ncert Exercises & Solutions

If the electronic configuration of an element is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}$ , the four electrons that participate in the chemical bond formation will be from :

1.	$3 p^{6}$	2.	$3 p^{6}, 4 s^{2}$
3.	$3 p^{6}, 3 d^{2}$	4.	$3 d^{2}, 4 s^{2}$

HINT: Electrons of ns and (n-1)d subshell take part in bond formation.

Explanation:

STEP 1:

The given electronic configuration shows that an element is a vanadium (Z = 22). It belongs to the d-block of the periodic table. In transition elements i.e., d-block elements, electrons of ns and (n-1)d subshell take part in bond formation.

STEP 2:

Hence, 3d and 4s electrons participate in bond formation. Therefore, option 4 is the correct answer.

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