Question

For an element, if electronic configuration is 1s22s22p63s23p63d34s2 then the element is

Answer 1

Explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}\mid}}}$

$1s {}^{2} \: , \: 2 {s}^{2} \: , \: 2 {p}^{6} \: , \: 3 {s}^{2} \: , \: 3 {p}^{6} \: , \: 4 {s}^{2} \: , \: 3 {d}^{3} \\ \\ atomic \: \: number \implies 23 \\ \\ 1 {s}^{2} \: \: \: \: 2 {s}^{2} \: \: \: \: \: \: \: 2 {p}^{6} \: \: \: \: \: \: \: \: \: \: \: 3 {s}^{2} \: \: \: \: \: \: \: \: 3 {p}^{6} \: \: \: \: \: \: \: \: \: \: \: \: 4 {s}^{2} \: \: \: \: \: \: \: 3 {d}^{3} \\ \boxed{\uparrow \downarrow} \: \boxed{\uparrow \downarrow } \: \boxed{\uparrow \downarrow} \boxed{\uparrow \downarrow} \boxed{\uparrow \downarrow } \: \boxed{\uparrow \downarrow } \: \boxed{\uparrow \downarrow} \boxed{\uparrow \downarrow} \boxed{\uparrow \downarrow } \: \boxed{\uparrow \downarrow} \: \: \boxed{\uparrow } \boxed{\uparrow } \boxed{\uparrow }$

We can easily located group just by observing electron number of d - orbital

it's 3..

So formula is

Group no.

= 2 + (no.of electron in d-subshell)

= 2 + 3

= 5

Group no. 5

Electron number 23

So atom is 'V' (Vanadium)