The Qstem says that the s orbital isn't hybridized, so that must mean the orbitals are (Px,py,pz) ->orthogonal to each other. With the absence of s orbitals, thus must mean orthogonal binding->90 degrees.
96 degrees is the closest to 90, so C

The Qstem says that the s orbital isn't hybridized, so that must mean the orbitals are (Px,py,pz) ->orthogonal to each other. With the absence of s orbitals, thus must mean orthogonal binding->90 degrees.
96 degrees is the closest to 90, so C

This is a bad question because it's a case of "the closest to 90" as opposed to "do you understand the idea?" The idea is that since bonding uses the p orbitals, which are orthogonal and oriented along the axes, the ligands must be orthogonal to each other. So you should be thinking "90" and if you can get to that point, you understand the idea behind the problem.

This is a bad question because it's a case of "the closest to 90" as opposed to "do you understand the idea?" The idea is that since bonding uses the p orbitals, which are orthogonal and oriented along the axes, the ligands must be orthogonal to each other. So you should be thinking "90" and if you can get to that point, you understand the idea behind the problem.

s orbital is spherically symmetric, so it will repel electrons in all directions equally and thus not affect the geometry about the metal.

That depends on the molecule. You have to be able to reason through each case. If you have unhybridized orbitals, then electrons in s-orbitals won't repel in any particular direction - that is, they repel in all directions symmetrically because the s-orbital is a sphere. However, if you have a similar case but with hybridized orbitals (as is the norm), then all electrons are in orbitals with equal energies that are not spherically symmetric (sp3 orbitals point towards the vertices of a tetrahedron) and therefore can repel in a particular direction.