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CBSE Questions for Class 12 Engineering Chemistry Coordination Compounds Quiz 13 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Coordination Compounds
Quiz 13
$$NCl_3$$ and $$BCl_3$$ are mixed in a container. Which the following is true for hybridisation of $$N$$ and $$B$$ the product formed by bonding of $$NCl_3$$ and $$BCl_3$$?
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Hybridisation of $$N$$ changes but for $$B$$ it remains same
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Hybridisation of $$B$$ changes but for $$N$$ it remains same
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Hybridisation of both $$N$$ and $$B$$ change
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Hybridisation of neither $$N$$ nor $$B$$ changes
The number of $${Cl}^{-}$$ combined by secondary valency in $$[Fe{({NH}_{3})}_{4}{Cl}_{2}]Cl$$ is ______
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1
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2
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3
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0
Which of the following complexes is not correctly matched with hybridization of its central metal?
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$$[CoF_6]^{3-} ; d^2 sp^3$$
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$$[Ni(CO)_4] ; sp^3$$
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$$[Cr (H_2O)_6]^{3+} ; d^2 sp^3$$
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$$[Pt (NH_3)_4]^{2+} ; d sp^2$$
Which compound to use for treatment of tumor?
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$$Cis\left[ Pd{ Cl }_{ 2 }{ \left( { NH }_{ 3 } \right) }_{ 2 } \right] $$
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$$Trans\left[ Pd{ Cl }_{ 2 }{ \left( { NH }_{ 3 } \right) }_{ 2 } \right] $$
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$$Cis\left[ Pt{ Cl }_{ 2 }{ \left( { NH }_{ 3 } \right) }_{ 2 } \right] $$
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$$Trans\left[ Pt{ Cl }_{ 2 }{ \left( { NH }_{ 3 } \right) }_{ 2 } \right] $$
Explanation
The given compound is cis plating which is used in the treatment of cancer or tumor.
Which of the following represent E isomer?
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Which of the following complexes form facial and meridional isomers?
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$$[Cr(H_2O)_5Cl]^{+2}$$
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$$[Cr(NH_3)_5Br]SO_4$$
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$$[Cr(NH_3)_3(NO_2)_3]$$
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$$[Cr(ox)_3]^{3-}$$
Tetravalency of carbon is possible in the case of
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$$ s p^{3} $$ -hybridisation
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$$s p^{2} $$ -hybridisation
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$$s p $$ -hybridisation
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all of these
The given compounds are______isomer of each other.
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Positional
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Chain
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Geometrical
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Functional
Which of the following statement is incorrect?
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The tendency to attract bonded pair of the electron in case of hybrid orbitals follows the order: $$sp > sp^2 > sp^3$$
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Alkali metals generally have negative value of electron gain enthalpy
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$$Cs^+(g)$$ releases more energy upon gain of an electron than $$Cl(g)$$
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The electronegativity values for $$2$$p-series elements is less than that for $$3$$p-series elements on account of small size and high inter electronic configuration
The compound $$PtCl_2\cdot 2NH_3$$ does not react with $$AgNO_3$$. This compound can be represented as?
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$$[Pt(NH_3)_2Cl]Cl$$
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$$[Pt(NH_3)_2Cl_2]$$
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$$[Pt(NH_3)_2]Cl_2$$
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both B and C
Which of the following represent $$sp - sp - sp^2 - sp - sp^2 - sp^3$$ hybridization from left to right order?
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$$CH_2= CH - C = \overset{H}{\overset{|}{C}} - C \equiv N$$
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$$CH \equiv C - CH = C = CH-CH_3$$
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$$N \equiv C - \overset{H}{\overset{|}{C}} - \underset{H}{\underset{|}{C}} - CH - CH_3$$
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Explanation
Carbon having single bond is $$sp^3$$ hybridised, carbon with double bond is $$sp^2$$ hybridised and the carbon having triple bond is $$sp$$ hybridised.
In option B $$CH$$ is $$sp$$ hybridised, $$C$$ is $$sp$$ hybridised, $$CH$$ is $$sp^2$$ hybridised, $$C$$ is $$sp$$ hybridised, $$CH$$ is $$sp2$$ hybridised and $$CH_3$$ is $$sp^3$$ hybridised.
In $$CuSO_4.5H_2O$$ how many molecules of water are indirectly connected to Cu?
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$$5$$
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$$4$$
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$$2$$
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$$1$$
Explanation
Solution:- (D) $$1$$
In $$CuSO_4.5H_2O$$, Four water molecules form coordinate bond with $$Cu_{2+}$$ ion while one water molecule is associated with $$H$$- bond
Which among the following is used in the treatment of cancer?
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cis-$$[Pt(en)_2Cl_2]$$
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cis-$$[PtCl_2(NH_3)_2]$$
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trans-$$[Pt(en)_2Cl_2]$$
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trans-$$[Pt(NH_3)_2Cl_2]$$
Explanation
$$cis-[PtCl_2(NH_3)_2]$$ also known as cisplatin is used as an anticancer drug in the chemotherapy process of cancer treatment.
Hence, option B is correct.
When two or more complexes are combined, it is
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Pattern
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Theme
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Complex
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All of these
The compound that inhibits the growth of tumors is:
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$$cis-[Pd(Cl)_2 (NH_3)_2]$$
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$$cis-[Pt(Cl)_2 (NH_3)_2]$$
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$$trans-[Pt(Cl)_2 (NH_3)_2]$$
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$$trans-[Pd(Cl)_2 (NH_3)_2]$$
Explanation
Solution:- (B) cis-$$\left[ Pt({Cl})_{2} {\left( N{H}_{3} \right)}_{2} \right]$$
$$cis-[Pt(Cl)_2 (NH_3)_2]$$ is used in chemotherapy to inhibits the growth of tumors.
The complex ion that will lose its crystal field stabilization energy upon oxidation of its metal to $$+3$$ state is:
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$$[Fe(phen)_3]^{2+}$$
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$$[Zn(phen)_3]^{2+}$$
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$$[Ni (phen)_3]^{2+}$$
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$$[Co(phen)_3]^{2+}$$
Explanation
By oxidation of $$Fe^{2+}$$ into $$Fe^{3+}$$, the CFSE value decrease.
(2) $$[Zn(phen)_3]_2\xrightarrow {-e^−}[Zn(phen)_3]^{3+}$$
$$Zn^{2+}$$: $$3d^{10}$$ $$Zn^{3+}$$: $$3d^9$$
C.F.S.E $$=0$$ C.F.S.E=$$−0.6\Delta _o$$
By oxidation of $$Zn^{2+}$$ into $$Zn^{3+}$$ , the CFSE value increase.
(3) $$[Ni(phen)_3]^{2+} \xrightarrow {-e^−}[Ni(phen)_3]^{3+}$$
$$Ni^{2+}$$: $$3d^8$$ $$Ni^{3+}$$: $$3d^7$$
C.F.S.E $$=−1.2{\Delta}_0$$ C.F.S.E $$=−1.8{\Delta}_0$$
By oxidation of $$Ni^{2+}$$ into $$Ni^{3+}$$ , the CFSE value increase.
(4) $$[Co(phen)_3]^{2+} \xrightarrow {-e^−}[Co(phen)_3]^{3+}$$
$$Co^{2+}$$: $$3d^7$$ $$Co^{3+}$$: $$3d^6$$
$$C.F.S.E=−1.8{\Delta}_0$$ $$C.F.S.E=−2.4{\Delta}_0$$
By oxidation of $$Co^{2+}$$ into $$Co^{3+}$$ , the CFSE value increase.
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale).
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Explanation
If both ligands present along z-axis removed from octahedral field and converted into a square planar field, then the energy level of orbitals is shown.
The degenerate orbitals of $${ \left[ Cr{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 3+ }$$ are:
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$${d}_{yz}$$ and $${d}_{{z}^{2}}$$
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$${d}_{{z}^{2}}$$ and $${x}_{xz}$$
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$${d}_{xz}$$ and $${d}_{yz}$$
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$${d}_{{x}^{2}-{y}^{2}}$$ and $${d}_{xy}$$
Explanation
Degenerate orbitals of $$[Cr(H_2O)_6]^{3+}$$ is shown in the diagram.
Hence according to the options given, degenerate orbitals are $$d_{xz}$$ and $$d_{yz}$$.
So, C option is correct.
Which complex among the following gives a white precipitate on treatment with an aqueous solution of barium chloride?
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$$[Pt(NH_3)_4Br_2]Cl_2$$
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$$[Co(NH_3)_5SO_4]NO_2$$
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$$[Co(NH_3)_5NO_2]SO_4$$
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$$[Pt(NH_3)_4Cl_2]Br_2$$
Explanation
White precipitate on treatment with barium chloride is due to the formation of
$$BaSO_4$$ and for this sulphate ion should be present in ionisable part.
$$[Co(NH_3)_5NO_2]SO_4 + BaCl_2(aq)\rightarrow [Co(NH_3)_5NO_2]Cl_2 + BaSO_4(s)\downarrow$$
$$BaSO_4$$ is a white precipitate.
The other compounds do not give white ppt. due to the absence of $$SO_4^{2-}$$ ions.
Hence, option $$C$$ is correct.
A molecule $$(X)$$ has (i) four sigma bonds formed by the overlap of $$sp^{2}$$ and $$s$$ orbitals; (ii) one sigma bond formed by $$sp^{2}$$ and $$sp^{2}$$ - orbitals and (iii) one $$\pi-bond$$ formed by $$p_{z}$$ and $$p_{z}$$ orbitals. Which of the following is $$X$$?
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$$C_{2}H_{6}$$
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$$C_{2}H_{5}Cl$$
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$$C_{2}H_{2}Cl_{2}$$
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$$C_{2}H_{4}$$
The complexes $$[CoNO_2(NH_3)_5]Cl_2$$ and $$[Co(ONO)(NH_3)_5]Cl_2$$ are not the examples of.
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Geometrical isomers
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Optical isomers
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Coordination isomers
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Linkage isomers
The total number of possible coordination isomer for the given compound $$[Pt(NH_3)_4Cl_2][PtCl_4]$$ is?
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$$2$$
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$$4$$
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$$5$$
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$$3$$
The octahedral complex $$CoSO_4Cl_5\,NH_3$$ exists in two isomeric forms $$X$$ and $$Y$$. Isomers $$X$$ reacts with $$AgNO_3$$ to give a white precipitate, but does not react with $$BaCl_2$$. Isomer $$Y$$ gives white precipitate with $$BaCl_2$$ but does not react with $$AgNO_3$$.
Isomers $$X$$ and $$Y$$ are:
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ionization isomers
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linkage isomers
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coordinating isomers
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solvent isomers
Explanation
$$\underset{\text{(X)}}{[Co(NH_3)_5SO_4]Cl}\xrightarrow{AgNO_3} \underset{\text{white ppt.}}{AgCl}\downarrow$$
$$\underset{\text{(Y)}}{[Co(NH_3)_5Cl]SO_4}\xrightarrow{BaCl_2}\underset{\text{white ppt.}}{BaSO_4}\downarrow$$
$$[X]$$ and $$[Y]$$ are ionization isomers.
Brown ring complex known as:
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$$K_4[Fe(CN)_6]$$
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$$[Fe(NO)(H_2O)_5]SO_4$$
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$$[Fe(NO)_2)(H_2O)_4]Br_2$$
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$$[FeF_6]^{3-}$$
Explanation
$$6FeSO_4+3H_2SO_4+2HNO_3\rightarrow 3Fe_2(SO_4)_3+4H_2O+2NO$$
$$FeSO_4+NO+5H_2O\rightarrow [Fe(NO)(H_2O)_5]SO_4$$
Brown ring complex
Option B is correct.
The complexes $$[Cr(NH_3)_6][Co(CN)_6]$$ and $$[Co(NH_3)_6][Cr(CN)_6]$$ are not the example of?
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Geometrical isomers
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Optical isomers
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Coordination isomers
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Linkage isomers
Select the correct statement(s) :
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$$ BF_3 > Me_3B $$
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$$ HC \equiv C-H (sp\ carbon) < H_2C = CH -H (sp^2 \ carbon)< H_3C- CH_2 -H(sp^3\ carbon ) $$
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$$ BF_3 < Me_3B $$
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$$ HC \equiv C -H (sp\ carbon) > H_2C=CH-H (sp^2\ carbon) > H_3C-CH_2 -H (sp^3\ carbon) $$
VSEPR theory suggest a small departure from $$ 109^0 28'$$ for $$ NH_3 $$ molecule. The H-N-H bond angle has been found to be $$ 107^0 $$ what kind of hybrid orbital will be occupied by the non -bonding pair of electron?
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$$ sp^3 $$
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$$ sp^{3.4} $$
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$$ sp^{2.13} $$
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$$ sp^2 $$
Select the correct statement for $$ P_4O_{10} $$
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It has four $$ sp^3 $$-hybridized phosphorous atoms
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It has higher $$s\%$$ charcter in $$P-O$$ bond than the $$ P_4O_6 $$
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It has cage -like structure
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It has $$ p_{ \pi} -d_{ \pi} $$ bonding
Explanation
From the above diagrams, we can conclude that :
All the four P atoms in P
₄O₁₀ are sp
³
hybridized (4 sigma bonds).
It has a cage-like structure.
The percentage s character in phosphorus pentoxide is higher than that in phosphorus trioxide due to the presence of double bond between P and O.
The bond between P atom and one of the O atoms of each PO
₄ unit in
P
₄O₁₀
is a double bond and involves p
π-dπ bonding.
Hence, all the above options are correct.
Select the correct options for following statement(s) :
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$$ sp^3 $$ hybrid orbitals are at $$ 90^0$$ to one another
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$$ sp^3d^2 $$ adjacent hybrid orbitals are at $$ 90^0 $$ to one another
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$$ sp^2 $$ hybrid orbitals are at $$ 120^0 $$ to one another
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Bond order of $$ N-O $$ bond in $$ NO_3^- $$ is $$ 1 \dfrac {1}{3} .$$
Choose the correct statement regarding the complexes X and Y:
(I) Both complexes have the same six geometrical isomer.
(II) Both complexes have nearly the same conductivity in $$H_2O$$.
(III) In both complexes, $$NH_3$$ is involved only to secondary valency.
(IV) In both complexes, electronic configuration of $$t_{2g}$$ is same.
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I, II and III
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I, II, III and IV
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II, III and IV
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I, III and IV
Which of the following sets of molecule(s) is/are having a linear shape but different hybridization?
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$$I^-_3$$ and $$CO_2$$
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$$HgCl_2$$ and $$ICl^-_2$$
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$$XeF_2$$ and $$C_2H_2$$
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$$XeF_2$$ and $$OCl^-$$
Explanation
A) $$I_{3}^{-} $$ and $$ CO_{2}$$ are linear in shape.
$$I_{3}^{-} $$ undergoes $$sp^{3} $$ hybridization and $$ CO_{2}$$ undergoes $$ sp $$ hybridization.
B) $$HgCl_{2} $$ and $$ICl_{2}^{-} $$ are linear in shape.
$$HgCl_{2} $$ undergoes $$ sp$$ hybridization and $$ICl_{2}^{-} $$ undergoes $$ sp^{3}d $$ hybridization .
C) $$XeF_{2} $$ and $$ C_{2}H_{2} $$ are linear in shape.
$$ XeF_{2} $$ undergoes $$sp^{3}d^{2} $$ and $$C_{2}H_{2} $$ undergoes $$ sp $$ hybridization.
D) $$XeF_{2} $$ and $$ OCl^{-} $$ are linear in shape.
$$ XeF_{2} $$ undergoes $$sp^{3}d^{2} $$ and $$ OCl^{-} $$ undergoes $$ sp^{3} $$ hybridization .
Hence, options A, B, C, and D all are correct.
Which of the following sets of the molecule(s) is/are having a V-shape but different hybridization?
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$$SnCl_2$$ and $$H_2O$$
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$$SO_2$$ and $$NO^+_2$$
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$$BF^-_2$$ and $$SCl_2$$
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$$OF_2$$ and $$SCl_2$$
In V.B.T., the idea of hybridization was required to explain which of the following facts:
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The equivalence of the bonds in most of the com-pounds
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The stereochemistry of the molecules
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The better overlapping of the orbitals
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None of these
Match the complex species given in Column-I with the possible isomerism given in Column-II and assign the correct code.
Column-I(Complex species)
Column-II(Isomerism)
(a) $$[Co(NH_3)_4Cl_2]^+$$
(p) Optical
(b) cis-$$[Co(en)_2Cl_2]^+$$
(q) Ionization
(c) $$[Co(NH_3)_5(NO_2)]Cl_2$$
(r) Coordination
(d) $$[Co(NH_3)_6][Cr(CN)_6]$$
(s) Geometrical
(t) Linkage
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a(p) b(q) c(s) d(t)
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a(s) b(r) c(q) d(p)
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a(s) b(p) c(q,t) d(r)
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a(s) b(p) c(q) d(r)
Complexes A and B are respectively.
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$$[Cr(NH_3)_4Cl_2]Br, [Cr(NH_3)_4Br]Cl_2$$
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$$[Cr(NH_3)_4Br]Cl_2, [Cr(NH_3)_4Cl_2]Br$$
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$$[Cr(NH_3)_4ClBr]Cl, [Cr(NH_3)_4Cl_2]Br$$
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$$[Cr(NH_3)_4Cl_2]Br, [Cr(NH_3)_4ClBr]Cl$$
Complex A is?
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$$[Cr(H_2O)_6]Cl_3$$
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$$[Cr(H_2O)_5Cl]Cl_2\cdot H_2O$$
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$$[Cr(H_2O)_4Cl_2]Cl\cdot 2H_2O$$
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None of these
One mole of the complex compound $$Co(NH_3)_5Cl_3$$ gives $$3$$ moles of ions on dissolution in water. One mole of the same complex reacts with two moles of $$AgNO_3$$ solution to yield two moles of $$AgCl(s)$$. The structure of the complex is?
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$$[Co(NH_3)_3Cl_3]\cdot 2NH_3$$
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$$[Co(NH_3)_4Cl_2]\cdot 2NH_3$$
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$$[Co(NH_3)_4Cl]Cl_2\cdot NH_3$$
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$$[Co(NH_3)_5Cl] Cl_2$$
Explanation
Total $$3$$ moles of ions are generated on dissolution in water, that is, one complex ion and two simple ions.
One mole of complex reacts with two moles of $$AgNO_3$$; thus, it must have $$2$$ Cl atoms bonded by ionic linkage. The coordination number of Co is $$6$$.
$$[Co(NH_3)_5Cl]Cl_2\to [Co(NH_3)_5Cl]^{2+}+2Cl^-$$
$$[Co(NH_3)_5Cl]Cl_2+2AgNO_3\to [Co(NH_3)_5Cl](NO_3)_2+2AgCl(ppt.)$$
Coordination compounds have great importance in biological systems. In this context, which of the following statements is incorrect?
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Cyanobalamin is $$B_{12}$$ and contains cobalt.
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Hemoglobin is the red pigment of blood and contains irons.
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Chlorophylls are green pigments in plants and contains calcium.
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Carboxypeptidase-A is an enzyme and contains zinc.
Explanation
The correct statement should be chlorophylls are green pigments in plants and contain magnesium.
The correct order regarding electronegativity of hybrid orbitals of carbon is:
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$$ sp < sp^2 < sp^3 $$
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$$ sp < sp^2 >sp^3 $$
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$$ sp > sp^2 < sp^3 $$
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$$ sp > sp^2 > sp^3 $$
Explanation
Electronegativity increases as s-character in hybrid orbital increases. Percentage of s character in $$ sp,\ sp^2$$ and $$sp^3 $$ are 50%, 33.33% and 25% respectively.
So, the correct option is D.
Carbon dioxide is isostructural with:
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$$HgCl_2$$
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$$SnCl_2$$
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$$C_2H_2$$
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$$NO_2$$
$$ 2Fe_2(SO_4)_3+3K_4[Fe(CN)_6]\rightarrow X+6K_2SO_4$$
What is the name of X? if-
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represents brown ring complex
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represents sodium nitroprusside
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represents Prussian blue
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represents chromyl chloride
What is the formula of X if
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represents $$CrO_{4}^{2-}$$
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represent $$CrO_{5}$$
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represent $$Cr_2O_{7}^{2-}$$
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represents $$CrO_2Cl_2$$
Explanation
$$\text{Option D is correct}$$
Here, X represents $$CrO_2Cl_2.$$
$$Na_2S+Na_2[Fe(CN)_5NO]\rightarrow X$$ (purple color)
What is the formula of X?
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$$Fe_4[Fe(CN)_6]$$
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$$[Fe(H_2O)_5NO]SO_4$$
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$$[Fe(H_2O)_5NO]$$
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$$Na_4[Fe(CN)_5(NOS)]$$
A solution containing $$2.675$$g of $$CoCl_3\cdot 6NH_3$$ (molar mass $$=267.5$$ g $$mol^{-1}$$) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $$AgNO_3$$ to give $$4.78$$g of $$AgCl$$(molar mass$$=143.5$$g $$mol^{-1}$$). The formula of the complex is?
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$$[Co(NH_3)_6]Cl_3$$
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$$[CoCl_2(NH_3)_4]Cl$$
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$$[CoCl_3(NH_3)_3]$$
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$$[CoCl(NH_3)_5]Cl_2$$
Explanation
$$AgCl$$ obtained is $$4.78$$g, that is, $$\dfrac{4.78}{143.5}$$ moles.
Thus, in $$\dfrac{2.675}{267.5}=10^{-2}$$ moles of complex, $$\dfrac{4.78}{143.5}\approx \dfrac{1}{30}$$ moles of $$Cl^-$$ ions were isonizable.
That is, in $$10^{-2}$$ moles of complex, $$\dfrac{1}{30}$$ moles $$Cl^-$$ are produced.
$$\therefore 1$$ mole of complex $$\rightarrow \dfrac{1}{30}\times 10^{2}$$ moles of $$Cl^-$$
$$=\dfrac{100}{30}=3.3$$ mols.
Thus, the formula of complex should be $$[Co(NH_3)_6]Cl_3$$.
Complex manure is that which supplies:
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$$S, K$$ and $$N$$
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$$N, K$$ and $$P$$
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$$S$$ and $$N$$
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$$S, N$$ and $$P$$
Explanation
Complex manures is that which supplies N, K and P that is nitrogen, potassium and phosphorus.
Nitrogen is good at making the leaves grow.
Phosphorus improves fruit and/or flower production as well as root growth.
Potassium is great for overall plant health.
Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given in the lists.
$$\{en =H_2NCH_2CH_2NH_2$$ 'atomic numbers; $$Ti=22; Cr=24; Co=27 Pt=78\}$$
List-I
List-II
(P) $$[Cr(NH_3)_4Cl_2]Cl$$
$$(1)$$ Paramagnetic and exhibits ionization isomerism
(Q) $$[Ti(H_2O)_5Cl](NO_3)_2$$
$$(2)$$ Dimagnetic and exhibits cis-trans isomerism
(R) $$[Pt(en)(NH_3)Cl]NO_3$$
$$(3)$$ Paramagnetic and exhibits cis-trans isomerism
(S) $$[Co(NH_3)_4(NO_3)_2]NO_3$$
$$(4)$$ Dimagnetic and exhibits ionization isomerism
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P-$$4$$, Q-$$2$$, R-$$3$$, S-$$1$$
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P-$$3$$, Q-$$1$$, R-$$4$$, S-$$2$$
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P-$$2$$, Q-$$1$$, R-$$3$$, S-$$4$$
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P-$$1$$, Q-$$3$$, R-$$4$$, S-$$2$$
Which one of the following has the largest number of isomers?
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$$[Ir(PR_3)_2H(CO)]^{2+}$$
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$$[Co(NH_3)_5Cl]^{2+}$$
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$$[Ru(NH_3)_4Cl_2]^+$$
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$$[Co(en)_2Cl_2]^+$$ ($$R=$$ alkyl group, $$en=$$ ethylenediamine)
What is the formula of brown ppt.?
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$$Cu_2I_2$$
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$$Cu_2I_2+I_{3}^{-}$$
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$$CuI_2$$
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$$CuSO_4$$
Which of the following complex is formed when A reacts with $$K_4[(Fe(CN)_6]$$?
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Prussian blue
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Turnbull's blue
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Brown ring complex
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Sodium nitroprusside
What is the formula of brown ppt?
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$$Fe(OH)_3$$
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$$Fe(OH)_2$$
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$$FeCl_3$$
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None of these
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