18. Of the following species, A) PH3 B) CIF3 C)NC13 D) BC3 B) All of these will

Question

18. Of the following species, A) PH3 B) CIF3 C)NC13 D) BC3 B) All of these will have bond angles of 120 will have bond angles of 120° 19. The molecular geometry of the left-most carbon atom in the molecule below is A) trigonal planar B) trigonal bipyramidal C) tetrahedral D) octahedral E) T-shaped 20·The central Xe atom in the XeF4 molecule has _ unbonded electron pair(s) and bonded electron pair(s) in its valence shell. A) 1,4 B) 2, 4 C) 4, 0 D) 4,1 E)4, 2 21. What is the molecular shape of H20? A) T-shaped B) tetrahedral C) linear D) trigonal pyramidal E) bent electron domains and a 22. PCIs has A) 6, trigonal bipyramidal B) 6, tetrahedral C) 5, square pyramidal D) 5, trigonal bipyramidal E) 6, scesaw

Explanation / Answer

18.

Answer is (D) BCl3

BCl3
Bonding pair = 3,
Lone pair = 0.
So it is a AX3 type molecule.

Hence bond angle = 120o

19.

Answer is (C) tetrahedral

The left most carbon is attached 3 hydrogens and 1 carbon. So, it is in sp3 hybridization. So it is tetrahedral geometry.

20.

Answer is (B) 2, 4

XeF4

8 + 4(7) = 36

Now for octet for four F to bonded to Xe = 8 x 4 = 32

So, 36 – 32 = 4 (that is two lone pair electrons)

In XeF4, four bonds, so 4 bond pair electrons.

21.

Answer is (E) bent (or V-shaped)

H2O
Bonding pair = 2,
Lone pair = 2.
So it is a AX2Y2 type molecule.

Hence shape is Non-linear (crooked) or bent or V-shaped.

22.

Answer is (D) 5, trigonal bipyramidal

P is less electronegative than Cl.
Thus, P becomes the central atom.
Thus 5 Cl will be the terminal atoms.
Bonding pair = 5,
Lone pair = 0.
So, it’s a AB5 type molecule

So, shape is trignonal bipyramidal

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