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p-Block Elements - I
normal
निम्नलिखित अणुओं में, $H - X - H$ कोण $( X = B , N$ अथवा $P )$ का क्रम होगा
A$PH _{3}\,< \,NH _{3}\,< \,NH _{4}^{+}\,< \,BF _{3}$
B$NH _{3}\,<\, PH _{3}\,<\, NH _{4}^{+}\,< \,BF _{3}$
C$BF _{3}\,< \,PH _{3}\,<\, NH _{4}^{+}\,<\, NH _{3}$
Dइनमे से कोई भी नहीं
(KVPY-2018)
Solution
(*) As the electronegativity of the central atom increases the bond angle also increases. So, $NH _{3}$ will have high bond angle than $PH _{3}$. The bond angle of $NH _{4}^{+}$is highest because of its symmetrical geometry, also bond pair and lone pair on $N$ of $NH _{3}$ suffer repulsion, which decreases its bond angle in comparison to $NH _{4}^{+}$.
Thus, the correct order of bond angle is
$PH _{3}\,<\, NH _{3}\,<\, NH _{4}^{+}$
But in the given question none of the option is correct because there is no $H – X – H$ bond angle in $BF _{3}$
Thus, the correct order of bond angle is
$PH _{3}\,<\, NH _{3}\,<\, NH _{4}^{+}$
But in the given question none of the option is correct because there is no $H – X – H$ bond angle in $BF _{3}$
Standard 11
Chemistry
