Question

Element Y with atomic number 3 combines with an element A having atomic number 17 . What would be the formula of the compound?

Answer 1

Element Y with atomic number 3 combines with an element A having atomic number 17.

So, we have an element whose electronic configuration is 3.

Therefore, in K shell it has 2 electrons and in L shell is has 1 electron.

→ K = 2 electrons and L = 1 electron (2+1 = 3)

So, to complete it's octet it will lose 1 electron and behave as a cation (positively charged ion.)

Similarly,

Electronic configuration of element A = 2 electrons in K shell, 8 electrons in L shell and 7 electrons in M shell.

So, to complete it's octet it will gain one electron and behave as an anion (negatively charged ion).

From above we have,

→ Y = +1 and A = -1

(Refer the attachment)

On cross-multiplying we get,

YA.

So, the compound is YA.

Answer 2

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$</p><p>\tt{\green{Element\: A \:has\: the\: atomic\: number \:17. }}$

Electronic configuration of element A :

=> 2 electrons in K shell, 8 electrons in L shell and 7 electrons in M shell.

So, to complete it's octet it will gain one electron and behave as an anion i.e, negatively charged ion.

$</p><p>\tt{\red{Element\: Y \:has\: the\: atomic\: number \:3. }}$

Therefore, in K shell it has 2 electrons and in L shell it has 1 electron.

→ K = 2 e- + L = 1 e- = 3 e-

To complete it's octet it will lose 1 electron and behave as a cation i.e, a positively charged ion.

$</p><p>\tt{\purple{Cross\: Y \: Multiplying, \: the\: compound\: becomes \:YA. }}$