Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Nitrates of group '1' (except LiNO_3) on heating give:

Answer»

`O_2`
`N_2`
NO
`NO_2`

ANSWER :A
Discussion
2.

Nitrates of all the metals are

Answer»

Coloured
Unstable
Soluble in WATER
Insoluble in water

SOLUTION :NITRATES and ACETATES of all METALS are water solution.
Discussion
3.

Nitrates of all metals are:

Answer»

Unstable
Coloured
Insoluble in water
Soluble in water

Answer :D
Discussion
4.

Nitrates of all metals are

Answer»

UNSTABLE
coloured
insoluble in WATER
SOLUBLE in water

Solution :Nitrates of all METALS are soluble in water
Discussion
5.

Nitrate is confirmed by ring test. The brown colour of the ring test is due to the formation of

Answer»

Ferrous NITRITE
`FeSO_(4).NO`
`FeSO_(4).NO_(2)`
Ferrous nitrate

Solution :`NaNO_(3)+H_(2)SO_(4)toNaHSO_(4)+HNO_(3)`
`6FeSO_(4)+3H_(2)SO_(4)+2HNO_(3)`
`to3Fe_(2)(SO_(4))_(3)+4H_(2)O +2NO`
`FeSO_(4)+NO to underset("(BROWN RING)")underset("Nitrosoferrous sulphate")(FeSO_(4).NO)`
Discussion
6.

Nitrate is confirmed by ring test. The brown colour of the ring is due to the formation of

Answer»

Ferrous nitrite
`FeSO_(4).NO`
`FeSO_(4).NO_2`
Ferrous nitrate

Solution :`FeSO_(4) +NO to FeSO_(4).NO` brown ring is due to COMPLEX `FeSO_(4).NO`
Discussion
7.

Nitrate of all metals are

Answer»

unstable
coloured
insoluble in water
soluble in water

Answer :D
Discussion
8.

Ni_((S)) |Ni_((aq))^(2+)||Ag_((aq))^(+)|Ag_((S)) is a non-standard cell, in which concentration of ion is less than 1 M, then determine correct formula to calculate potential of non-standard cell.

Answer»

`E_(cell)=E_(cell)^(THETA)-(0.059)/(2)"log"([A_(G)^(+)])/([Ni^(2+)])`
`E_(cell)=E_(cell)^(Theta)-(0.059)/(2)"log"([Ni^(2+)])/([AG^(+)])`
`E_(cell)=E_(cell)^(Theta)-(0.059)/(2)"log"([Ni^(2+)]^(2))/([Ag^(+)]^(2))`
`E_(cell)=E_(cell)^(Theta)-(0.059)/(2)"log"([Ni^(2+)])/([Ag^(+)]^(2))`

Solution :The following cell reaction is occurred
`Ni_((S)) + 2Ag_((aq))^(+) to Ni_((aq))^(2+)+2Ag_((S))"Where, "n=2`
`K=(["Product"])/(["Reactant"])=([Ni_((aq))^(2+)][Ag_((S))]^(2))/([Ni_((S))][Ag_((aq))^(+)]^(2))=([Ni_((aq))^(2+)])/([Ag_((aq))^(+)]^(2))`
So, `E_(cell)=E_(cell)^(theta)-(0.059)/(2)"log"([Ni_((aq))^(2+)])/([Ag_((aq))^(+)]^(2))`.
Discussion
9.

Nitrate & Nitrite both give brown ring test, can be distinguish by -

Answer»

`HOSO_(2)NH_(2)` (Sulphonic acid)
`NH_(2)HgO.Hgl` ("Million base")
`FeSO_(4)`
None

Answer :A
Discussion
10.

nipropyl cyanideon reductiongives

Answer»

n-butyl AMINE
n-propyl amine
ISOBUTYLAMINE
t-butyl amine

ANSWER :A
Discussion
11.

Niobium crystallizes in body centred cubic structure. If its density is 8.55 g cm^(-3), calculate the atomic radius of niobium. (Atomic mass of Nb = 93u , N_A = 6.02 xx 10^(23) mol^(-1))

Answer»

Solution :DENSITY of niobium, d=8.55 G `cm^(-3)`, Number of atoms in bcc UNIT cell, Z= 2
Atomic mass of niobium, M = 93u
Density, `d = (ZM)/(N_A xx a^3)`
`a^3=(ZM)/(d xx N_A)=(2 xx 93)/(8.55 xx 6.02 xx 10^(23))`
`= 3.61 xx 10^(-23) cm^3= 36.1 xx 10^(-24) cm^3`
`a = 3.30 xx 10^(-8) cm=3.30 Å`
For a bcc unit cell, `4r = a SQRT3` or `r = (a sqrt3)/(4)=(3.30 xx sqrt3)/(4) Å`
= `(3.30 xx 1.73)/(4) = 1.427` Å= 14.27 nm
Discussion
12.

Niobium crystallizes in body-centred cubic structure. If its density is 8.55 g cm^(-3), calculate atomic radius of niobium, given its atomic mass 93 u.

Answer»

SOLUTION :`a^(3) = (M xx Z)/(d xx N_(0) xx 10^(-30))`
`a^(3) = (93 G mol^(-1) xx 2)/(8.55 g cm^(-3) xx 6.02 xx 10^(23) mol^(-1) xx 10^(-30))`
`a^(3) = 3.61 xx 10^(7) = 36.1 xx 10^(6)`
`a = (36.1)^(1//3) xx 10^(2)` pm
`= 3.304 xx 10^(2)` pm = 330.4 pm
`[x = (36.1)^(1//3), log x = 1//3 log 36.1`
`= 1//3 xx 1.5575 = 0.519` x
= antilog 0.519 = 3.305]
For body-centred cubic, `r = (SQRT(3))/(4) a = 0.4333`a
`= 0.433 xx 330.4` pm = 143.1 pm
Discussion
13.

Niobium crystallises as body centered cube (BCC) and has density of 8*55kg//dm^(3). Calculate the atomic radius of niobium. (Given : Atomic mass of niobium = 93). Write one statement of first law of thermodynamics and its mathematical expression. Write the reactions involved in the zone of reduction in blest furnace during extraction of iron.

Answer»

SOLUTION :Density of Niobium (Nb) crystal is `=8*55kg//dm^(3)`
Crystalline structure is bec.
Atomic mass no. of Nb =93 g/mol.
Avogadro number `=N_(A)=6*022xx10^(23)mol^(-1)`
Atomic radius of Niobium = ? In bcc unit cells, there are 8 atoms at 8 -coners and 1 atoms at body centre.
`:.` Number of Nb atoms. `=(1)/(8)xx8+1=1+1=2`
Mass of one Nb atom `=(a^(2))/(6*022xx10^(23))`
`=1*5544xx10^(-22)`
`=3*088xx10^(-22)kg`
Mass of unit cell = Mass of 2 Nb atoms
`=4*088xx10^(-22)`g
If a is edge LENGTH of bcc unit cell Volume of unit cell `=^(3)`
Density `=("Mass of unit cell")/("Volume of unit cell")`
`d=(3*088xx10^(-22))/(a^(3))`
`:.""a^(3)=(3*088xx10^(-22))/(d)`
`=(3*088xx10^(-22))/(8.55)`
`=0*361xx10^(-22)dm^(3)`
`=36*1xx10^(-24)dm^(3)`
`a=(36*1xx1-^(-24))^(1//3)`
`=3*33xx10^(-8)dm`
If r is the radius of 1 Nb atom. then in bcc structure
`r=(sqrt(3))/(4)a`
`=(sqrt(3))/(4)xx3*33xx10^(-8)`
`=1*43xx10^(-8)dm`
`or""~~14*3nm`
(where, 1 dm `=10^(-9)` nm)
Hence, the atomic radius of niobium is `14*3` nm.
The first law of thermodynamics states that Energy can NEITHER be created nor be destroyed. The total energy of an isolated system (cannot exchange energy with its surroundings) remains the same.
Mathematical expression :
In a system that exchanges both heat and mechanical work with its surroundings, the total energy or the energy change is described by taking both - heat and work in it's account.
`U_(2)-U_(1)=DeltaY=q+W`
`q=DeltaU-W`
`DeltaU=q+W`
This is the mathematical expression for first law of thermodyamics, from the point of view of conservation of energy. This expression justifies first law of thermodynamics.
where,
`U_(2)andU_(1)` =Energies of system in the final and initial states, respectively
`DeltaU` =Change in internal energy
q=heat absorbed or evolved by the system
W = Work DONE by the system or the system.
Reactions involved in the of reduction :
`3Fe_(2)O_(3)+COto2Fe_(3)O_(4)+CO_(2)`
`Fe_(3)O_(4)+COto3FeO+CO_(2)`
`FeO+COtoFe+CO_(2)`.
Discussion
14.

NiO has been found to dissolve substantial amount of Li_(2)O. As Li^(+) occupies some cation sites, which among the following can be inferred from the above statement?

Answer»

NIO CONTAINS of `Ni^(2+), LI^(+)` and `O^(2-)` as the only ions.
NiO has equal number of `Li^(+)` ions and `Ni^(2+)` ions
NiO has equal number of `Li^(+)` ions and `Ni^(3+)` ions.
NiO has equal number of `Ni^(2+)` and `Ni^(3+)` ions.

ANSWER :C
Discussion
15.

Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm^(3), calculate atomic radius of niobium using its atomic mass of 93 u.

Answer»

Solution :The equation CONNECTING these quantities is given by
`a^(3)=(Mxxz)/(dxxN_(A))`,
Substituting the values, we get
`a^(3)=(93xx2)/(8.55xx6.022xx10^(23)) or a=330.4xx10^(-10)cm or a=330.4" PM"`
For body - centred cubic, `r=(SQRT3)/(4)a=0.433xx330.4=143.1" pm."`
Discussion
16.

Niobium crystallises in body-centred cubic structure. If density is 8.55 "g cm"^(-3), calculate atomic radius of niobium using its atomic mass 93 u.

Answer»

Solution :Density (d) = `8.55 "G cm"^(-3)`
Atomic radius (r) = ?
Number of atoms per unit cell (Z) = 2
By formula `d = (Z XX M)/(N_A xx a^3) ` we get
`a^3 = (2 xx 93)/(8.55 xx 6.022 xx 10^(23))`
`therefore a= 3.304 xx 10^(-8)` cm
`therefore a = 330.4 "pm"`
For BCC, `r = (SQRT(3))/(4) a`
`r = (sqrt3)/(4) xx (330.4)` = 143.1 pm.
Discussion
17.

NiO is antiferromagnetic. But on heating at 250^(@)C, it becomes paramagnetic. Why?

Answer»

Solution :In solid state, the magnetic moments of NiO are aligned in such a FASHION, such that the net moment is ZERO. So. NiO is ANTIFERROMAGNETIC. On heating, the alignment is DISTURBED. The domains are randomized.`Ni^(2+)` has `d^(8)` configuration with unpaired electrons, hence, exhibits paramagnetism
Discussion
18.

[Ni(NH_(3))_(2)]^(2+)overset(Conc.HCl)(to)A+B The molecular formula of both A and B is same.A can be converted to B by boiling in dil. HCl. A on reaction with oxalic acid yields a complex having the formula Ni(NH_(3))_(2)(C_(2)O_(4)) but B does not. From the above information we can say that

Answer»

A is square planar but B is tetrahedral
A and B both are tetrahedral A is optically active compound WHEREAS B is optically inactive
both A and B are square planar A is transisomer and B is CIS-isomer
both A and B are square planar A is cis-isomer and B is trans -isomer.

Solution :`[Ni(NH_(3))_(2)]^(2+)overset(CONC. HCl)(to) [Ni(NH_(3))_(2)Cl_(2)]+[Ni(NH_(3))_(2)Cl_(2)]`
In this complex the oxidation state of Ni is +2

In complex `[Ni(NH_3)_2Cl_2],dsp^2`

THUS the complex is square planar
The two isomers of this complex are

The cis isomer can easily form a chelate ring with oxalate group.
The cis -ismers can be converted to trans isomer on boiling with dil HCl.
Hence A is cis -isomer and B is trans -isomers.
Discussion
19.

Nine volumes of gaseous: mixture consisting of gaseous organic, compound A and just sufficient amount of O_(2) required for complete combustion yielded on burning 4 volumes of CO_(2), 6 volumes of water vapour and 2 volumes of N_(2) all volumes measurêd at the same temperature and pressure. If the compound A contained only C, H and N (i) How many volumes of O_(2) required for complete combustion ? (ii) What is the molecular formula of the compound A?

Answer»

SOLUTION :`C_(2)H_(6)N_(2)`
Discussion
20.

Ni(II) compounds are thermodynamically more stable than Pt(II)- Justify your answer.

Answer»

Solution :(i) The IONISATION enthalpy values can be used to predict the THERMODYNAMIC STABILITY of their compounds.
(ii) `{:("For Nickel,"I.E_(1)+I.E_(2),=,737+1753),(,=,2490" kJ mol"^(-1)),("For Platinum,"I.E_(1)+I.E_(2),=,864+1791),(,=,"2655 kJ mol"^(-1)):}`
Since, the energy required to FORM `NI^(2+)` is less than that of `Pt^(2+), Ni(II)` compounds are thermodynamically more stable than `Pt(II)` compounds.
Discussion
21.

[Ni(H_(2)O_(6)]^(2+) absorbs….

Answer»

GREEN light
Red light
Yellow light
Orange light

Answer :B
Discussion
22.

Night blindness is due to the deficiency of

Answer»

VITAMIN A
vitamin B
vitamin C
vitamin D

Answer :A
Discussion
23.

Night blindness is caused by deficiency of vitamin :

Answer»

VIT-`B_12`
Vit-A
Vit-C
Vit-E

Solution : NIGHT blindness is CAUSED by DEFICIENCY of VITAMIN A.
Discussion
24.

[Ni(CO)_(x)] and [Fe(CO)_(y)] what will be the value of x and y respectively are :

Answer»

4 and 4
4 and 5
5 and 4
5 and 5

Solution :`[Ni(CO)_(4)]rArrEAN=36, [Fe(CO)_(5)]rArrEAN=36.`
Discussion
25.

Night blindness and kertinisation of skin is due to the deficiency of …….. .

Answer»

VITAMIN`B_(1)`
vitamin C
Vitamin A
vitamin`B_(12)`

Answer :C
Discussion
26.

Nicotinamide is named for

Answer»

VIT-`B_2`
vit-`B_5`
vit-`B_6`
vit-`B_(12)`

ANSWER :B
Discussion
27.

Ni(CO)_4 is an example of

Answer»

cationic COMPLEX
NEUTRAL complex
ANIONIC complex
poly nuclear complex

Answer :B
Discussion
28.

[Ni(CN)_4 ]^(2-)is diamagnetic ,while[NiCl_4]^(2) -is paramagnetic ,explain using crystal field theory.

Answer»

Solution :(a) `[Ni(CN)_(4)]^(2-)`
`Ni=3d^(8)4S^(2)`
`Ni^(2+)=3d^(8)`

Nature of the complex - Low spin (Spin paired)
Ligand filled electronic configuration of central metal ion, `t_(2g)^(6)e_(g)^(2)`
Magnetic PROPERTY : No unpaired ELECTRON (`CN^(-)` is strong filled ligand), HENCE it is diamagnetic.
Magnetic moment : `mu_(s)=0`
(b) `[NiCl_(4)]^(2-)`
`Ni=3d^(8)4S^(2)`
`Ni^(2+)=3d^(8)`

Nature of the complex, - high spin
Ligand filled electronic configuration of central metal ion, `t_(2g)^(6)e_(g)^(2)`
magnetic property : Two unpaired electron (`Cl^(-)` is weak field ligand). Hence it is paramagnetic
Magnetic moment : `mu_(s)=sqrt(2(2+2))`
`=sqrt(8)=2.83BM`
Discussion
29.

[Ni(CN)_(4)]^(2-) is diamagnetic , while [NiCl_(4)]^(2-) is paramagentic, explain using crystal field theory.

Answer»

Solution :`(a) [Ni(CN)_(4)]^(2-)`
`Ni=3d^(8)4S^(2)`
`Ni^(2+)=3d^(8)`

Nature of the complex-Low spin (Spin PAIRED)
Ligand filled electronic configuration of central metal ion, `t_(2g)^(6)e_(g)^(2)`
Magnetic property : No unpaired electron (`CN^(-)` is strong filled ligand ), hence it is dimagnetic.
Magnetic MOMENT : `mu_(s)=0`
`(b)=[NiCl_(4)]^(2-)`
`Ni=3d^(8)4S^(2)`
`Ni^(2+)=3d^(8)`

Naute of the complex,-HIGH spin
Ligand filled electronic configuration of central metal ion , `t_(2g)^(6)e_(g)^(2)`
Magnetic property : Two unpaired electron (`Cl^(-)` is weak field ligand). Hence it is paramagnetic
Magnetic moment : `mus=sqrt(2(2+2))`
`=sqrt(8)=2.83BM`
Discussion
30.

[Ni(CN)_(4)]^(2-), [MnBr_(4)]^(2-)and [FeF_(6)]^(3-). Geometry, hybridisation and magnetle moment of the ions respectively are

Answer»

Tetrahedral ,square planar, octabedral: `sp^(3), DSP^2,sp^(3)d^(2):5.9,0,4.9`
Tetrahedral ,square planar, OCTAHEDRAL: `dsp^(2), sp^(3),sp^(3)d^(2) : 0,5.9, 4.9`
Square planar, tetrahedral ,octahedral: `dsp^2, sp^3, d^2sp^3 :5,9,4.9,0`
Square planar, tetrahedral ,octahedral: `dsp^2, sp^3, sp^2 d^2: 0,5.9,4.9`

Answer :D
Discussion
31.

[Ni(CN)_(4)]^(2-) is colourless whereas [Ni(H_(2)O)_(6))^(2+) is green. Why?

Answer»

SOLUTION :In `[Ni(H_(2)O)_(6)]^(2+), Ni^(2+)` has TWO unpaired electrons which do not pair up in the presence of WEAK ligand `H_(2)O`, THEREFORE, excitation from lower d-orbitals to higher d-orbitals gives the colour.
Discussion
32.

[Ni(CN)_(4)]^(2-) is colourless whereas [Ni(H_(2)O)_(6)]^(2+) is green. Why ?

Answer»

Solution :`H_(2)O` is a weak LIGAND. 3d electrons of `Ni^(2+)` do not pair up. The unpaired electrons absorb red wavelength of white light for `d-d` transition and transmitted colour (complementary colour) is green. HENCE, the solution of `[Ni(H_(2)O)_(6)]^(2+)`is green.
`CN^(-)` is a strong ligand. The 3d electrons of `Ni^(2+)` pair up. There is no unpaired ELECTRON left to absorb ENERGY. Hence, solution of `[Ni(CN)_(4)]^(2+)` is colourless.
Discussion
33.

[Ni(CN)_(4)]^(2-) complex has the shape …… and is ….. In terms of magnetic nature whereas [NiCl_(4)]^(2-) has the shape …… and is ….. .

Answer»

Solution :SQUARE PLANAR, DIAMAGNETIC , TETRAHEDRAL, paramagnetic
Discussion
34.

[Ni(CN)_(4)]^(2-) and [NiCl_(4)]^(2-) have similarity but not in :

Answer»

MAGNETIC MOMENT
C.N. and O.N.
geometry
both (a) and ©

Answer :D
Discussion
35.

Explain on the basis of V.B, theory that [Ni(CN)_4]^(2-) ion with square planar structure is diamagnetic and the [NiCl_4]^2- ion with tetrahedral geometry is paramagnetic.

Answer»

SOLUTION :`[NI(CN)_4]^2-`
Discussion
36.

[Ni(CN)_4]^(2-) has square planar structure and it is diamagnetic. Identify the ligand in the mentioned complex

Answer»

SOLUTION :CYANIDE ION
Discussion
37.

[Ni(CN)_(4)]^(2-)" and "[NiCl_(4)]^(2-) have similarity but not in :

Answer»

MAGNETIC MOMENT
C.N. and O.N.
structure
both (a) and (c )

ANSWER :D
Discussion
38.

[NiCl_(4)]^(2-) is paramagnetic while [Ni(CO)_(4)] is dimagnetic though both are tetrahedral. Why?

Answer»

Solution :Ni is in +2 oxidation state in `[NiCl_(4)]^(2-)` with the configuration `3d^(8)4s^(0).Cl^(-)` is weak LIGAND. It cannot pair up the electrons in 3d ORBITALS. Hence, it is PARAMAGNETIC. In `[Ni(CO)_(4)]`, Ni is in ZERO oxidation state with the configuration `3d^(8) 4s^(2)`. In the presence of strong CO ligand the 4s electrons SHIFT to 3d to pair up with 3d electrons. Thus, there is no unpaired electron left. Hence, it is diamagnetic.
Discussion
39.

Nicl_(4)]^(2-) is paramagnetic while [Ni(CO_(4))] is diamagnetic though both are tetrahedral. Why

Answer»

Solution :In `[Nicl_(4)]^(2-)`, `Ni.is +2` oxidation STATE with the CONFIGURATION `3d^(8) 4s^(@)`. Cr is weak ligand. It cannot pair up the ELECTRONS in 3d orbitals. Hence, it is paramagnetic. In `[Ni(CO)_(4)]` Ni is in ZERO oxidation state with the configuration `3d^(8) 4s^(2)` In the presence of CO ligand, the 4s electrons shift to 3d to pair up 3d electrons. Thus, there is no UNPAIRED electron present. Hence, it is diamagnetic.
Discussion
40.

[NiCl_(4)]^(2-) is paramagnetic while [Ni(CO)_(4)] is diamagnetic though both are tetrahedral. Why ?

Answer»

Solution :In `[NiCl_(4)]^(2-)`, Ni is +2 oxidation STATE with the configuration `3d^(8)4s^(0)` . `CL^(-)` is weak ligand. It cannot pair up the electrons in 3d orbitals. Hence, it is paramagnetic. In `[Ni(CO)_(4)]`, Ni is in zero oxidation state with the configuration `3d^(8)4s^(2)`. In the presence of `CO` ligand, the 4s electrons shift to 3d to pair up 3d electrons. Thus, there is no unpaired electron PRESENT. Hence, it is DIAMAGNETIC.
Discussion
41.

[NiCl_(4)]^(2-) has geometry

Answer»

SQUARE planar
tetrahedral
square bipyramidal
trigonal bipyramidal

Answer :A::B::D
Discussion
42.

[NiCl_(2){P(C_(2)H_(5))_(2)(C_(6)H_(5))]_(2)] exibits temperature dependent magnetic behaviour (paramagnetic/diamangetic). The coordination geometries of Ni^(2+) in the paramagnetic and diamagnetic states are respectively

Answer»

tetrahedral and tetrahedral
SQUARE planar and square planar
tetrahedral and square planar
square planar and tetrahedral

Solution :In BOND states (paramagnetic and diamagnetic) of the GIVEN complex, Ni exists as `Ni^(2+)` whose electronic configuration is `[AR]3d^(8)4s^(0)`

In the above paramagnetic state the geometry of the complex is `sp^(3)` giving tetrahedral geometry.
the diamagnetic state is achieved by PAIRING of electrons in 3d orbital.

Thus the geometry of the complex will be `dsp^(2)` giving square planar geometry.
Discussion
43.

[NiCl_(4)]^(2-) is paramagnetic while [Ni(CO)_(4)] is diamagnetic though both are tetrahedral. Why?

Answer»

Solution :In`[NiCl_(4)]^(-2)`, Ni has `3d^(8)4s(0)` CONFIGURATION, `CL^(-)` can.t pair up while in`[Ni(CO)_(4)]`, Ni has`3d^(8)4s(2)`configuration, CO pair uo ELECTRONS.
Discussion
44.

NiCl_(2)overset(KCN)underset(HCN) to "Complex " A NiCl_(2)overset(KCN)underset(excess) to "Complex " B A and B complexes have the co-ordination number 4. The hybridization of both complexes are :

Answer»

`dsp^(2)`
`SP^(2)" and "dsp^(2)`
`dsp^(2)" and "sp^(3)`
both `sp^(3)`

Solution :`A to underset(dsp^(2)("diamag".))(K_(2)[Ni(CN)_(4)]),""B to underset(sp^(3)(2 u.p.e^(-)))(K_(2)[NiCl_(4)])`
Discussion
45.

NiCl_(2)overset(KCN)underset(HCN) to "Complex " A NiCl_(2)overset(KCN)underset(excess) to "Complex " B A and B complexes have the co-ordination number 4. What are the magnetic nature of A and B ?

Answer»

Both diamagnetic.
A is diamagnetic and B is paramagnetic with ONE unpaired electrons.
A is diamagnetic and B is paramagnetic with two unpaired electrons.
Both are paramagnetic.

Solution :`A to underset(dsp^(2)("diamag".))(K_(2)[NI(CN)_(4)]),""B to underset(sp^(3)(2 u.p.e^(-)))(K_(2)[NiCl_(4)])`
Discussion
46.

NiCl_(2)overset(KCN)underset(HCN) to "Complex " A NiCl_(2)overset(KCN)underset(excess) to "Complex " B A and B complexes have the co-ordination number 4. The IUPAC name of complexes A and B are respectively :

Answer»

Potassium tetracyanonickelate(II) and Potassium tetrachloronickelate(II)
Potassium tetracyanonickel (II) and Potassium tetrachloronickel(II)
Potassium cyanonickelate(II) and Potassium chloronickel(II)
Potassium cyanonickel(II) and Potassium chloronickel(II)

Solution :`A to UNDERSET(DSP^(2)("diamag".))(K_(2)[Ni(CN)_(4)]),""B to underset(sp^(3)(2 u.p.e^(-)))(K_(2)[NiCl_(4)])`
Discussion
47.

NiCl_(2) {P(C_(2)H_(5))_(2)(C_(6)H_(5))}_(2) exhibits temperature dependent magnetic behaviour (paramagnetic/diamagnetic). The coordination geometries of Ni^(2+) in the paramagnetic and diamagnetic states are respectively

Answer»

tetrahedral and tetrahedral
square planar and square planar
tetrahedral and square planar
square planar and tetrahedral

Solution :`[NiCl_(2){P(C_(2)H_(5))_(2)(C_(6)H_(5))}_(2)]`
contains `Ni^(2+)` having the electronic configuration as under :

In the high spin state, it is paramagnetic with `sp^(3)` hybridization giving tetrahedral geometry. In the LOW spin state (unpaired electrons PAIR up), it is diamagnetic with `dsp^(2)` hybridization giving square planar geometry.
Discussion
48.

NiCl_(2) overset(KCN)to "complex"A NiCl_(2) underset("excess") overset(KCl)to "complex"B A & B complexes have the co-ordination number 4. What are the magnetic nature of 'A' & 'B' ?

Answer»

Both DIAMAGNETIC
A' is diamagnetic & 'B' is PARAMAGNETIC with one UNPAIRED ELECTRONS.
A' is diamagnetic & 'B' is paramagnetic with two unpaired electrons.
Both are paragmagnetic

Answer :C
Discussion
49.

NiCl_(2) overset(KCN)to "complex"A NiCl_(2) underset("excess") overset(KCl)to "complex"B A & B complexes have the co-ordination number 4. The hybridisation of both complexes are :

Answer»

`DSP^(2)`
`SP^(2) & dsp^(2)`
`dsp^(2) & sp^(3)`
both `sp^(3)`

Answer :C
Discussion
50.

NiCl_(2) overset(KCN)to "complex"A NiCl_(2) underset("excess") overset(KCl)to "complex"B A & B complexes have the co-ordination number 4. The IUPAC name of complexes 'A' & 'B' are respectively :

Answer»

POTASSIUM tetracyanonickelate(II) and Potassium tetrachloronickelate(II)
Potassium tetracyanonickel(II) and Potassium tetrachloronickel(II)
Potassium cyanonickelate(II) and Potassium chloronickelate(II)
Potassium cyanonickel(II) and Potassium chloronickel(II)

ANSWER :A
Discussion