The ground state electronic configuration of S is 3s2 3p4. How does it form the compound SF6?

Question

The ground state electronic configuration of S is 3s2 3p4. How does it form the compound SF6? (A) Due to octahedral shape of S atoms (B) Due to presence of vacant 3d -orbitals which provide 6 unpaired electrons in excited state (C) Due to sp3 hybridisation of S atom which provides 6 electrons to 6 F atoms (D) Due to presence of 3 sigma and 3 pi bonds between S and F

Tardigrade Admin · Accepted Answer

S can go into excited state with 6 unpaired electrons due to presence of vacant 3d - orbitals, which are overlapped by six F electrons. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/60835559544cad689cef7e3f537de00e-.png" />

Q. The ground state electronic configuration of $S$ is $3 s^{2} 3 p^{4}$ . How does it form the compound $S F_{6}$ ?

Due to octahedral shape of $S$ atoms

Due to presence of vacant $3 d$ -orbitals which provide $6$ unpaired electrons in excited state

Due to $s p^{3}$ hybridisation of $S$ atom which provides $6$ electrons to $6 F$ atoms

Due to presence of $3$ sigma and $3$ pi bonds between $S$ and $F$

$S$ can go into excited state with $6$ unpaired electrons due to presence of vacant $3 d$ - orbitals, which are overlapped by six $F$ electrons.

Q. The ground state electronic configuration of S is 3s23p4. How does it form the compound SF6​?

Due to octahedral shape of S atoms

Due to presence of vacant 3d -orbitals which provide 6 unpaired electrons in excited state

Due to sp3 hybridisation of S atom which provides 6 electrons to 6F atoms

Due to presence of 3 sigma and 3 pi bonds between S and F