Chemical bonding

Q.4) well,in NH₃,there are lonepair-bondedpair and bondedpair-bondedpair repulsions,which slightly distorts its bondangle to just less than 109^o,while in H₂O,THERE IS L.P.-L.P.REPULSION ,which is more effective than two repulsions in NH₃,which brings down its angle to about 104^o.now,in OF₂,becoz the electrons are nearer to flourine atom due to high electronegativity,so its distorted to maximum.and finally in Cl₂O, the bonded electron pairs are closer to oxygen making repulsion between them more and thereby reducing the l.p.-l.p. repulsion on oxygen.also due to bulkiness of Cl,the bond angle increases to about 111^o.
thus,Cl₂O has greater bond angle

do we ever consider whether the bonded atoms have lone pairs or not in deciding structure. We only care for lone pair of central atom

Consider SO₂ do you consider that O has lone pairs or do u just see how many lone pairs or bond pairs the S (central atom) has??

@maydayhay

yes steric factor is the main reason here..

electronegativity order is F>O>N>Cl>>>>>H

so the other reason given by him shld apply to H2O as well..as u said

Q3) it depends on orientation of d orbitals..(no need to worry at this stage)
Q2)(c)

@yes: can you explain Q2 and Q1. how do you know its (c)??

symmetry factor of the octahedral structure and the two orbitals

wats the meaning of this?

Q.1) In an octahedral complex,the metal is at the centre and the ligands are at six corners

. It follows this way that the approach of six ligands along the x,-x,y,-y,z,-z directions will increase the energy of the e_g orbitals (d_{x² - y²},d_z²) (which point along the axes) ,much more than the t_2g orbitals(d_xy,d_xz,d_yz),(which point between the axes).

thus a) is the answer.