Question

1.Element A, B, C, D have atomic numbers 17, 8, 10, 11 respectively. Which one

a) will form a cation b) will form an anion c) is a metal d) is a noble gas​

Answer 1

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To answer all this questions we should first write the electronic configuration of this element.

Element A =

Atomic number = 17

Therefore, electronic configuration is

k shell - 2

L shell - 8

M shell - 7

still this element requires one more electron to make itself stable.

so this element is a cation

Element B =

Atomic number = 8

Therefore, electronic configuration is

k shell - 2

L shell - 6

still this element requires two more electron to make itself stable .

so this element is a cation .

Element C =

Atomic number = 10

Therefore, electronic configuration is

K shell - 2

L shell - 8

The outermost shell of this element is completely filled . This makes the element stable .

so this element is a noble gas.

Element D =

Atomic number= 11

k shell - 2

L shell - 8

M shell - 1

The outermost shell of this element contains one electron but still it requires 7 electrons to make itself stable but it can't gain 7 electrons so it loses the 1 electron and becomes stable .

so this element contains one extra electron

That is why Element D is anion .

your answers:

a. Element A and B

b. Element D

c. Element D

d. Element C

Note :

If an element needs electrons then it is cation.

If an element loses electrons then it is anion.

Hope this is useful...