d2sp3 hybridisation of the atomic orbitals gives
square planar structure
triangular structure
tetrahedral structure
octahedral structure
A group of atoms can function as a ligand only when
It is a small molecule
It has an unshared electron pair
Its a negatively charged ion
It is a positively charge ion
The pair of compounds having identical shapes for their molecules is
CH4, SF4
BCl3, ClF3
XeF2, ZnCl2
SO2, CO2
The correct arrangement for the ions in the increasing order of their radii is
Na+, Cl-, Ca2+
Ca2+, K+, S2-
Na+, Al3+, Be2+
Cl-, F-, S2-
B.
Ca2+, K+, S2-
Since, Ca2+, K+, S2- have same number of electrons so they are isoelectronic ions, as we know that, the ion having greater number of proton will be smallest in size. Thus, the increasing order of ionic radii is Ca2+ < K+ < S2-.
The correct arrangement of the species in the decreasing order of the bond length between carbon and oxygen in them is
CO, CO2, HCO, CO
CO2,CO, CO, HCHO
CO, HCO, CO2, CO
CO, CO, CO2, HCO
A correct statement is
[Co(NH3)6]2+ is paramagnetic
[MnBr4]2- is tetrahedral
[CoBr2(en)2]- exhibits linkage isomerism
[Ni(NH3)6]2+ is an inner orbital complex
Which one of the following conversion results in the change of hybridisation and geometry?
CH4 to C2H2
NH3 to N+H4
BF3 to BF
H2O to H3O+
If the bond energies of H-H, Br-Br and H-Br are 433, 192 and 364 kJ mol-1 respectively, then H° for the reaction-
H2 (g) + Br2 (g) → 2HBr (g) is
-261 kJ
+103 kJ
+261 kJ
-103 kJ