Coordination Compounds Test

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Coordination Compounds Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

Primary and secondary valency of Pt in $$[Pt{ (en) }_{ 2 }{ Cl }_{ 2 }]{ Cl }_{ 2 }$$ are:

A
4,4

B
2,6

C
6,6

D
None of these

Solution

The Primary valency is the number of negative ions that are present as counter ions to the complex.

The secondary valency is the number of ligands that are attached to the metal ion. Hence the Primary valency is 2 and the secondary valency is 6.

Therefore option B is correct.
Question 2
1 / -0

Which of the following complex will give white precipitate with BaCl$$_2$$ (aq)?

A
[Co(NH$$_2)_4SO_4]NO_2$$

B
[Cr(NH$$_3)_5SO_4$$]Cl

C
[Cr(NH$$_3)_5Cl]SO_4$$

D
Both $$B$$ and $$C$$

Solution

Since the reagent is $$BaCl_2(aq)$$, displacement reaction will take place. The resulting solution, if insoluble in water, will form a white precipitate. Both $$BaNO_2$$ and $$BaCl_2$$ are soluble in water and hence will not form a precipitate. $$BaSO_4$$, however, is insoluble in water and hence forms a white precipitate. Therefore C is the correct answer
Question 3
1 / -0

Which of the molecules is not tetrahedral?

A
$$\left[ Pt(en)_{ 2 } \right] ^{ 2+ }$$

B
$$\left[ Ni(Co)_{ 4 } \right]$$

C
$$\left[ Zn(NH_{ 3 })_{ 4 } \right] $$

D
$$\left[ { NiCl }_{ 4 } \right] ^{ 2- }$$

Solution

The Complex $$[Pt(en)_2]^{2+}$$ is not tetrahedral because in this complex the metal ion is Pt which is a present in 5d series and the 5d series element have more diffused orbitals than 3d series elements and hence the ligand have the more closest approach towards metal ion and therefore the complex formed is inner orbital square planar. Hence the correct option is A.
Question 4
1 / -0

Which of the following statements is true about hybridization ?

A
Hybrid orbitals frequently undergo linear overlaps making sigma bonds.

B
Hybrid orbitals frequently undergo lateral overlaps maing $$\pi$$-bonds. In other words, there are several compounds in which $$\pi$$-bonds are formed using hybrid orbitals.

C
Hybrid orbitals are molecular orbitals.

D
A hybrid orbital bigger in size makes shorter bond.

Solution

$$(A)$$
Linear overlaps of bybridabitak from sigma bonds
$$True$$
$$(B)$$
HyBrid orbitals Do Not Form $$\pi$$-bonds. $$\pi$$-bonds are formed by arbybridized orbitals of the molecule by lateral overlap.
Statement is $$False$$
$$(C)$$
Hybrid are not same as molecules orbital hence $$False$$ statement
While of same atom, molecular orbitals are formed by mixing orbitals of different atoms
$$False$$
$$(D)$$
Only first Statement is $$True$$
Question 5
1 / -0

Which of the following name is impossible ?

A
Potassium tetrafluoridooxidochromate($$VI$$)

B
Barium tetrafluoridobromate($$III$$)

C
Dichlorobis(urea)copper($$II$$)

D
All are impossible
Question 6
1 / -0

In 2sp hybridization, 2s-orbital can be mixed with

A
Only $$2p_x$$

B
Only $$2p_y$$

C
Only $$2p_z$$

D
Any one of $$2p_x,2p_y$$ and $$2p_z$$

Solution

$$S$$ orbital is spherically symmetric and hence can be mixed with $$p$$ orbital from any axis
So, answer should be $$d$$
Question 7
1 / -0

Hybridization of sulphur in $$SF$$$$_{4}$$ molecules are :

A
$$sp$$$$^{3}d^{3}$$

B
$$sp$$$$^{3}$$d

C
$$sp$$$$^{3}d^{2}$$

D
$$dsp$$$$^{2}$$

Solution
Question 8
1 / -0

The bond angle and hybridization in ether $$(CH_3OCH_3)$$ is ______.

A
$$106.51^0 , sp^{2}$$

B
$$110^0$$ , $$sp^3$$

C
$$109.28^0, sp^3$$

D
None of these.

Solution
Question 9
1 / -0

Choose the correct options:

A
In $$O_2F_2$$, the $$O-F$$ bond length is lesser than expected value.

B
In trislyl amine $$(SiH_3)_3$$N, the bond length is lesser than expected value

C
The bond angle in $$OF_2$$ is lesser than in $$Cl_2O$$

D
The Be atom in $$BeCl_2(s)$$ is $$sp^3$$ hybridised.

Solution
Question 10
1 / -0

Copper sulphate when dissolved in excess of $$KCN$$ gives

A
$$Cu(CN)_2$$

B
$$[Cu(CN)_4]^{3-}$$

C
$$[Cu(CN)_4]^{2-}$$

D
$$CuCN$$

Solution