The electronegativity of the following elements increases in the order
A.
$$C,N,Si,P$$
B.
$$N,Si,C,P$$
C.
$$Si,P,C,N$$
D.
$$P,Si,N,C$$
Answer :
$$Si,P,C,N$$
Solution :
NOTE: Electronegativity increases on moving from left to right in a period and decreases on moving from top to bottom in a group.
$$Si$$ and $$P$$ are placed in the $${3^{{\text{rd}}}}$$ period while $$C$$ and $$N$$ are placed in the $${2^{{\text{nd}}}}$$ period. Elements in $${2^{{\text{nd}}}}$$ period have higher electronegativities than those in the $${3^{{\text{rd}}}}$$ period. Since $$N$$ has smaller size and higher nuclear charge than $$C,$$ its electronegativity is higher than that of $$C.$$ Similarly, the electronegativity of $$P$$ is higher than that of $$Si$$ . Thus, the overall order is : $$Si,P,C,N.$$
Releted MCQ Question on Inorganic Chemistry >> Classification of Elements and Periodicity in Properties
Releted Question 1
The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is