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27151.

The number of Fe^(3+) ions in one molecular of prussian blue is.

Answer»


ANSWER :4
Discussion
27152.

The number of vacant orbitals of element with atomic number 14 is:

Answer»

2
4
8
6

Answer :D
Discussion
27153.

The number of FIF angles which are less than 90^@ in IF_5​ :-

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6
Zero
8
4

Solution :
Discussion
27154.

The number of Faradays needed to reduce 4 gram equivalents ofCu^(++)to Cu metal will be

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1
2
`1//2`
4

Solution :NUMBER of GM equivalent=number of FARADAY PASS
4gm=4faraday.
Discussion
27155.

The number of vacant hybrid orbitals which participate in the formation of 3-centre 2 electrons bonds i.e., banana bonds in diborane structure is :

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SOLUTION :
No of vacant `sp^3` hybrid ORBITALS participating in the formation of banana bonds is 2.
Discussion
27156.

The number of Faradays of electricity required to decompose 100 ml water (density = 0.99 gm/ml) is

Answer»

2
11
100
5.5

Answer :B
Discussion
27157.

The number of Faradays needed to reduce 4 gm. Equivalents of Cu^(2+) to Cu metal will be

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1
2
`1/2`
4

Solution :`CU^(2+)+2E^(-) TOCU`
Discussion
27158.

The number of unpaired in the central metal ion of brown ring complex is

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Solution :`Fe^(+) = [Ar]3d^(6)4S^(1)` with `SP^(3)d` hybridization . 4s ELECTRON paired with 3d
Discussion
27159.

The number of Faradays needed to reduce 4 gm equivalents of Cu^(2+) to Cu metal will be

Answer»

1
2
`1//2`
4

Solution :1 G eq.require 1 F.
Discussion
27160.

The number of unpaired electrons present in Cr^(3+)is:

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3
1
2
5

Answer :A
Discussion
27161.

The number of Faradays needed to reduce 4 g equivalents ofCu^(2+)to Cu metal will be

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1
2
`1//2`
8

Answer :D
Discussion
27162.

The number of unpaired electrons present in complex ion [FeF_(6)]^(3-) is:

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5
4
6
0

Solution :`[FeF_(6)]^(3-) to d^(5)"(HIGH SPIN)"`
Discussion
27163.

The number of faraday required to liberate 1 mole of any element indicates :

Answer»

WEIGHT ELEMENT
CONDUCTANCE of electrolyte
Charge on the ION of that element
None

Answer :C
Discussion
27164.

The number of unpaired electrons present in Am are

Answer»

2
3
4
7

Answer :B
Discussion
27165.

Thenumber of Faraday required to gneerate 1g of Mg from MgCl_2 is:

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1
2
3
4

Answer :B
Discussion
27166.

The number of unpaired electrons in Yb^(3+) is found to be

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ZERO
ONE
two
six

Answer :B
Discussion
27167.

The number of ether metamers represented by theformulaC_(4) H_(10)is -

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4
3
2
1

Answer :B
Discussion
27168.

The number of unpaired electrons ind^(6), low spin, octahedral complex is :

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4
2
1
0

Answer :D
Discussion
27169.

The number of ether metamers represented by molecular formula C_(4)H_(10)O is

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4
3
2
1

Solution :The three metamers having `C_(4)H_(10)O` as the molecular formula are (i) `C_(2)H_(5)O(C_(2))_(5)`,
(II) `CH_(3)OCH_(2)CH_(2)CH_(3)` (iv) `CH_(3)OCH(CH_(3))_(3)`
Discussion
27170.

The number of unpaired electrons in the square planar [Pt(CN)_(4)]^(2-) ion is

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2
1
0
3

Solution :In the given compled `[Pt(CN)_(4)]^(2-)`
The electron configuration of Pt
`Pt to [Xe]4f^(14)5D^(9)6s^(1)`
`Pt^(2+)to[Xe]4f^(14)5d^(8)`

In has zero UNPAIRED electron as `CN^(-)` is a strong field ligand.
Discussion
27171.

The number of ether isomes for molecular formula C_(4)H_(10)O is ________

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2
3
4
5

Answer :B
Discussion
27172.

The number of essential amino acids in man is

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8
10
18
20

Solution :The AMINO ACIDS which can't be synthesised by HUMAN body so they are essential to take from diet. They are 10 in NUMBER
Discussion
27173.

The number of unpaired electrons in the square planar complex [Pt(CN)_4]^(2-) is

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2
3
0
1

Answer :C
Discussion
27174.

The numberof essentialamino acidin man is …… .

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8
10
20
19

Answer :B
Discussion
27175.

The number of unpaired electrons in the brown ring complex is

Answer»


SOLUTION :
Discussion
27176.

The number of unpaired electrons in [NiCl_(4)]^(2-), Ni(CO)_(4) and [Cu(NH_(3))_(4)]^(2+) respectively are

Answer»

2,2,1
2,0,1
0,2,1
2,2,0

Solution :In `[NiCl_(4)]^(2-)`, Ni is in +2 oxidation STATE.

Unpaired electrons =2
In `Ni(CO)_(4),Ni` is in zero oxidation state.
`Ni:(AR]3d^(8)4s^(2)`

Co is STRONG field ligand and causes pairing of electron of 4s to 3d.
`Ni(CO)_(4)`

Unpaired electron =0
In `[Cu(NH_(3))_(4)]^(2+)`, Cu is in +2 oxidation state.
`Cu^(2+): [Ar]3d^(9)4s^(0)`
`[Cu(NH_(3))_(4)]^(2+)`

Unpaired ELECRTRON =1
Discussion
27177.

The number of equivalent Cr-O bonds in CrO_4^(2-) is

Answer»

1
2
3
4

Answer :D
Discussion
27178.

The numberofenantiomeric paristhatcan beproduct duringmonochlorination Of 2-mehtybutane is :

Answer»

2
3
4
1

Answer :A
Discussion
27179.

The number of unpaired electrons in outer orbital [Fe(H_(2)O)_(6)]^(3+) complexis :

Answer»

1
3
5
2

Answer :C
Discussion
27180.

The number of elements that can be accommodated in the present set up of the long form of the periodic table is (upto 7^(th) period).

Answer»

`117`
`118`
`119`
`120`

ANSWER :B
Discussion
27181.

The number of elliptical orbits, including circular orbits in the M shell of an atom is:

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3
4
2
1

Answer :A
Discussion
27182.

The number of unpaired electrons in Ni^(3+). is

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3
2
4
8

Answer :B
Discussion
27183.

The number of elements in the fifth period of periodic table is

Answer»

8
10
18
32

Answer :C
Discussion
27184.

The number of elements in each of the long periods in the periodic table is

Answer»

2
8
18
32

Answer :C
Discussion
27185.

The number of unpaired electrons in Ni (atomic number = 28) are

Answer»

0
2
4
8

Solution :`Ni:(28)to[Ar]:`
`THEREFORE` No. of unpaired ELECTRONS= 2
Discussion
27186.

The number of elements in each long period is :

Answer»

18
10
8
32

Answer :A
Discussion
27187.

The number of unpaired electrons in Ni^(2+) is :

Answer»

Zero
2
4
8

Answer :B
Discussion
27188.

The number of elements in each of the inner transition series is

Answer»

2
8
10
14

Solution :Lanthanides- The `14` elements form `._(58)Ce-_(71)Lu` in which 4F subshell is being PROGRESSIVELY filled up are called lanthanides or rare earth elements . Actinides-Similarly, the `14` elements from `._(90)TH-_(103)LR` in which 5f-subshell is being progressively filled up are called actinides.
Discussion
27189.

The number of unpaired electrons in Mn^+is:

Answer»

3
5
4
6

Answer :D
Discussion
27190.

The number of electrons to balance the following equation, the value of x is -

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5
4
3
2

Answer :C
Discussion
27191.

The number of unpaired electrons in Lu ^(3+) are

Answer»

0
2
6
7

Answer :A
Discussion
27192.

The number of electrons that have a total charge of 9650 coulombs is

Answer»

`6.22 TIMES 10^(23)`
`6.022 times 10^(24)`
`6.022 times 10^(22)`
`6.022 times 10^(-34)`

Answer :C
Discussion
27193.

The number of unpaired electrons in [Fe(CN)_(6)]^(3-) is.............and the magnetic moment value is

Answer»


ANSWER :`1,1.73BM`
Discussion
27194.

The number of electrons that have a total charge of 9650 coulombs is …………………. .

Answer»

`6.22 xx 10^(23)`
`6.022 xx 10^(24)`
`6.022 xx 10^(22)`
`6.022 xx 10^(-34)`

SOLUTION :IF = `96500C = 1 "mole of " e^(-) = 6.023 xx 10^(23) e^(-)`
`:. 9650 C = (6.22 xx 10^(23))/(96500) = 6.022 xx 10^(22)`.
Discussion
27195.

The number of unpaired electrons in [Fe(CN)_(6)]^(3-) is :

Answer»

three
ONE
four
SIX.

Answer :B
Discussion
27196.

The number of electrons that have a total charge of 965 coulombs is

Answer»

`6.022xx10^(23)`
`6.022xx10^(22)`
`6.022xx10^(21)`
`3.011xx10^(23)`

ANSWER :A::B::C
Discussion
27197.

The number of electrons required to reduced4.5 xx 10^(-5)g of Al is

Answer»

` 1.03xx10^(18)`
`3.01 XX 10^(18)`
`4.95 xx 10^(26)`
`7.31 xx 10^(20)`

Solution :`underset(27)(AL^(3+)) +3e^(-) to underset(27g)(Al)`
27 g of Al is reduced by `=3 xx 6.023xx10^(23)E^(-)s`
` 4.5xx10^(-5)` g of Al will be reduced by
`=(3xx6.023xx10^(23) xx 4.5xx10^(-5))/(27)`
`=3.01 xx 10^(18)` electrons
Discussion
27198.

The numberof electronsrequiredto balancethe followingequation, NO_(3)^(-) +4H^(+) +e^(-)to2H_2 O +NOis

Answer»

5
4
3
2

Solution :`MO_(3)^(-)+4H^(+)to 2H_2 O +NO`
in thisequationall theatomsarebalanced.To balancechargeAdd`3E^(-)`to L.H.Swe have
`NO_3^(-)+ 4H^(+)+ 3e^(-)to 2H_2O +NO `
Discussion
27199.

The number of unpaired electrons in d^(7), low spin, octahedral complex is

Answer»


SOLUTION :In low spin octahedral complexes `Delta_(0)gtP`. Hence the electronic arrangement will be:`t_(2g)^(6)e_(G)^(1)`
Discussion
27200.

The number of unpaired electrons in d^(6), low spin, octahedral complex is :

Answer»

4
2
1
0

Solution :`d^(6) to t_(2)^(2,2,2)e_(G)^(0,0)`
Discussion