Question

# Which of the following statement(s) is/are correct?$$S_1: SeF_6$$ has the octahedral structure and $$PF_5$$ has $$sp^3d^2$$ hybridization.$$S_2$$: The percentage of s-character in the orbital forming S−S bonds and P−P bonds in $$S_8$$ and $$P_4$$ molecules respectively are 25% and 33.3%.$$S_3:$$ $$[Co(NH_{3})_{6}]^{3+}$$ is an inner orbital complex as well as diamagnetic in behaviour.

A
Only S1
B
S2 and S3
C
S1 and S3
D
Only S3

Solution

## The correct option is D Only $$S_3$$$$S_1:$$ The central Se atom has 6 bond pairs of electrons and 0 lone pairs of electrons. Itundergoes $$sp^3d^2$$ hybridisation, which results in octahedral geometry. In case of $$PF_5$$, thecentral P atom has 5 sets of electron pairs (all of which are bonding) and $$sp^3d^2$$ hybridizationcorresponds to 6 sets of electron pairs.Therefore $$PF_5$$ is $$sp^3d$$ hybridized. Hence, statement 1 isfalse.$$S_2: S_8(sp^3d^2)$$ and $$P_4(sp^3)$$ both has $$25\%$$ s- character.$$S_3:$$ Orbitals in $$Co^{+3}$$ ion$$\boxed {\uparrow \downarrow} \boxed {\uparrow} \boxed {\uparrow}\boxed {\uparrow}\boxed{\uparrow}$$ $$\boxed { }$$ $$\boxed { } \boxed { } \boxed { }$$$$3d$$ $$4s$$ $$4p$$Making Hybridized orbitals$$\boxed {\uparrow \downarrow} \boxed {\uparrow \downarrow} \boxed {\uparrow\downarrow}$$ $$\boxed { } \boxed { } \boxed { } \boxed { } \boxed { }$$$$d^2sp^3$$$$[Co(NH_3)_6]^{3+}$$$$\boxed {\uparrow \downarrow} \boxed {\uparrow \downarrow} \boxed {\uparrow\downarrow}$$ $$\boxed {\uparrow \downarrow}\boxed {\uparrow \downarrow}\boxed {\uparrow\downarrow}\boxed {\uparrow \downarrow}\boxed {\uparrow \downarrow}\boxed {\uparrow\downarrow}$$$$6$$ pairs of electrons from $$6 \ NH_3$$ molecules.Inner '$$d$$' orbitals are used in hybridization so its inner orbital complex. There are nounpaired electrons so that it is diamagnetic.Option D is correct.Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More