Question

The electronic configuration of the following elements K${}^{19}$, Mn${}^{25}$, Ca${}^{20}$ are given.

K${}^{19}$ = 1s${}^2$ 2s${}^2$ 2p${}^6$ 3s${}^2$ 3p${}^6$ 4s${}^x$

Mn${}^{25}$ = 1s${}^2$ 2s${}^2$ 2p${}^6$ 3s${}^2$ 3p${}^6$ 3d${}^5$ 4s${}^y$

Ca${}^{20}$ = 1s${}^2$ 2s${}^2$ 2p${}^6$ 3s${}^2$ 3p${}^6$ 4s${}^z$

The product of x, y, and z is:

Answer 1

The electronic configurations of elements are as given below.

K${}^{19}$ = 1s${}^2$ 2s${}^2$ 2p${}^6$ 3s${}^2$ 3p${}^6$ 4s${}^1$

Hence, $x=1$.

Mn${}^{25}$ = 1s${}^2$ 2s${}^2$ 2p${}^6$ 3s${}^2$ 3p${}^6$ 3d${}^5$ 4s${}^2$

So, $y=2$.

Ca${}^{20}$ = 1s${}^2$ 2s${}^2$ 2p${}^6$ 3s${}^2$ 3p${}^6$ 4s${}^2$

Hence, $z=2$

Then the product of x, y and z is $1\times 2\times 2=4$