From the following bond energies:
H-H bond energy: 431.37 kJ mol-
C=C bond energy: 606.10 kJ mol-
C- C bond energy: 336.49 kJ mol-
C-H bond energy: 410.50 kJ mol-
Enthalpy for the reaction,
will be
1523.6 kJ mol-
-243.6 kJ mol-
-1200. kJ mol-
-1200. kJ mol-
Which one of the element with the following outer orbital configuration may exhibit the largest number of oxidation states?
3d5,4s2
3d5,4s1
3d5,4s2
3d5,4s2
C.
3d5,4s2
A number of oxidation states exhibited by d-block elements is the sum of the number of electrons (unpaired) in d-orbitals and number of electrons in s - orbital.
a) 3d3, 4s2 ⇒ Oxidation state = 3+2 = 5
b) 3d5, 4s1 ⇒ Oxidation state = 5+1 = 6
c) 3d5, 4s2 ⇒ Oxidation state = 5+2 = 7
d) 3d4, 4s2 ⇒ Oxidation state = 2+ 2 = 4
Hence, an element with 3d5, 4s2 configuration exhibits the largest number of oxidation states.
The ionisation constant of ammonium hydroxide is 1.77 x 10-5 at 298 K. Hydrolysis constant of ammonium chloride is
5.65 x 10-10
6.50 x 10-12
5.65 x 10-13
5.65 x 10-13
Oxidation number of P in PO43-, of S in SO42- and that of Cr in Cr2O72- are respectively,
+5,+6 and +6
+3, + 6 and +5
+5,+3 and +6
+5,+3 and +6
Amongst the elements with following electronic configuration, which one of them may have the highest ionisation energy?
[Ne] 3s2 3p3
[Ne] 3s2 3p2
[Ar] 3d10, 4s2 4p3
[Ar] 3d10, 4s2 4p3
Maximum number of electrons in a subshell of an atom is determined by the following
4l + 2
2l +1
4l-2
4l-2
The values of ΔH and ΔS for the reaction, C(graphite) + CO2(g) → 2 CO (g) are 170 kJ and 170 JK-1 respectively. This reaction will be spontaneous at
710 K
910 K
1110 K
1110 K
Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO3 and Conc. H2SO4.In the mixture, nitric acid acts as a/an:
reducing agent
acid
base
base
According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order?