Sub-shell electronic configuration of elements P, Q and R is given.
[P, Q and R are imaginary symbols]
P $= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}$ ,
Q $= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}$ ,
R $= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2} 4 s^{2} 3 d^{1}$
(a) One of these electronic configurations given above is wrong. Which is the element?
(b) Out of P, Q and R which has the highest electronegativity?
(c) Can P and Q join to form an ionic compound? Justify your answer.

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Solution

(a) The electronic configuration of the element R is wrongly given. This is because the electron filling in the subsequent sub-shell will take place only after filling 6 electrons in the 3p sub-shell.
(b) Element Q has the highest electronegativity.
(c) P and Q can join to form an ionic compound. There is only one electron $(4 s^{1})$ in the outermost shell of the element P. So, it can easily donate that electron to Q. Q can accept that electron as it has only 7 electrons in the outermost shell and attain the octet form. Thus, an ionic compound PQ is formed by the transfer of electrons.

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L (l s^{2}, 2 S^{2}, 2 p^{4}); P (l s^{2}, 2 s^{2}, 2 P^{6}, 3 s^{1})

Q (l s^{2}, 2 s^{2} 2 P^{6}, 3 s^{2} 3 p^{5})

R (l s^{2}, 2 s^{2} 2 P^{6}, 3 s^{2})

There are four elements P, Q, R and S with their electronic configuration given in the options. Which of these elements belong to the same group?

(1 s^{2}, 2 s^{2} 2 p^{4}); Q (1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{5}), p (1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{1}); R (1 s^{2} 2 s^{6} 2 p^{6}, 3 s^{2})

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