Question

Write the electronic configuration of 9 F
19
, 16 S
32 and 18 Ar 38
and then point out the element with

a) maximum nuclear charge
b) minimum number of neutron
c)max. number of unpaired electron​

Answer 1

9^F^19 = 1s^2 2s^2 2p^2x 2p^2y 2p^1z

16^S^32 = 1s^2 2s^2 2p^6 3s^2 3p^2x 3p^1y 3p^1z

18^Ar^38 = 1s^2 2s^2 2p^6 3s^2 3p^6  

(i) Max. nuclear charge =18 in .18^Ar^38

(ii) Minimum no. of neutrons =10 in .9^F^19

(iii) Maximum no. of unpaired electrons =2 in .16^S^32

Hope it helps...

Answer 2

Answer:

a) ${}^{38}_{18} Ar$

b) ${}^{19}_9 F$

c) ${}^{32}_{16}S$

Given:

${}^{19}_9 F$ , ${}^{32}_{16}S$, ${}^{38}_{18} Ar$ are three given elements.

Find:

The electronic configuration of the elements given.

Solution:

1. ${}^{19}_9 F$  (Fluorine)

Atomic number, Z = 9

Mass number, A = 19

Number of electrons = Number of protons = Z = 9

Number of neutrons = A - Z = 19 - 9 = 10

Nuclear Charge = Number of protons×e = +9e

Electronic configuration = 1s² 2s² 2p⁵

Unpaired electrons = 1

2. ${}^{32}_{16} S$ (Sulphur)

Atomic number, Z = 16

Mass number, A = 32

Number of electrons = Number of protons = Z = 16

Number of neutrons = A - Z = 32-16 = 16

Nuclear Charge = Number of protons×e = +16e

Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁴

Unpaired electrons = 2

3. ${}^{38}_{18} Ar$ (Argon)

Atomic number, Z = 18

Mass number, A = 38

Number of electrons = Number of protons = Z = 18

Number of neutrons = A - Z = 38 - 18 = 20

Nuclear Charge = Number of protons×e = +18e

Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶

Unpaired electrons = 0

a) Ar has the maximum nuclear charge. (+18e)

b) F has the minimum number of neutrons. (10)

c) S has the maximum number of unpaired electrons. (2)

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