Q: The correct order of the wavelength maxima of the absorption band in the ultraviolet-visible region for the given complexes is

_ O 1 aligned & The absorb wave length order is \\ & [ Co ( CN )_6 ]^ 3- < [ Co ( NH _3 )_6 ]^ 3+ < [ Co ( NH _3 )_5 H _2 O ]^ 3+ < [ Co ( NH _3 )_5 Cl ]^ 2+ aligned

Q: The sum of the spin only magnetic moment values (in B.M.) of [ Mn ( Br )_6 ]^ 3- and [ Mn ( CN )_6 ]^ 3- is _____.

Sum of spin magnetic moments of both complexes is 7.70 to 7.73 ~B . M .

Q: Among [ Co ( CN )_4 ]^ 4- , [ Co ( CO )_3( NO ) ], XeF _4, [ PCl _4 ]^ + , [ PdCl _4 ]^ 2- , [ ICl _4 ]^ - , [ Cu ( CN )_4 ]^ 3- and P _4 the total number of species with tetrahedral geometry is _____

[ Co ( CN )_4 ]^ 4- , [ Co ( CO )_3 NO ], [ PCl _4 ]^ + , [ Cu ( CN )_4 ]^ 3- ~\&~ P _4 are with tetrahedral geometry.

Q: Among the following options, select the option in which each complex in Set-I shows geometrical isomerism and the two complexes in Set-II are ionization isomers of each other. [en .= H _2 NCH _2 CH _2

Set-I: $ [ Co ( NH _3 )_3 ( NO _2 )_3 ]$ and $ [ Co (e n)_2 Cl _2 ]$ Set-II: $ [ Co ( NH _3 )_5 Cl ] SO _4$ and $ [ Co ( NH _3 )_5 ( SO _4 ) ] Cl $ Set-I: Set-II: $ [ Co ( NH _3 )_5 Cl ] SO _4$ and $ [ Co ( NH _3 )_5 SO _4 ] Cl $ are ionisation isomers.

Q: Among V ( CO )_6, Cr ( CO )_5, Cu ( CO )_3, Mn ( CO )_5, Fe ( CO )_5, [ Co ( CO )_3 ]^ 3- , [ Cr ( CO )_4 ]^ 4- , and Ir ( CO )_3, the total number of species isoelectronic with Ni ( CO )_4 is _______

In case of complexes, isoelectronic species should be those having same effective atomic number (EAN) Ni ( CO )_4 28+4 2=36 (i) V ( CO )_6 23+2 6=35 (ii) Cr ( CO )_5 24+2 5=34 (iii) Cu ( CO )_3 29+2 3=35 (iv) Mn ( CO )_5 25+2 5=35 (v) Fe ( CO )_5 26+2 5=36 (vi) [ Co ( CO )_3 ]^ 3- 27+3+2 3=36 (vii) [ Cr ( CO )_4 ]^ 4- 24+4+2 4=36 (viii) [ Ir ( CO )_3 ] 77+2 3=83

Q: Among the following complexes, the total number of diamagnetic species is_____ [ Mn ( NH _3 )_6 ]^ 3+ , [ MnCl _6 ]^ 3- , [ FeF _6 ]^ 3- , [ CoF _6 ]^ 3- , [ Fe ( NH _3 )_6 ]^ 3+ , and [ Co ( en )_3 ]

Mn ^ 3+ [ Ar ] 3 ~d ^4 d ^4 configuration in t _ 2 ~g and eg _ g orbitals will always have unpaired electrons irrespective of SFL and WFL. Fe ^ 3+ [ Ar ] 3 ~d ^5 d ^5 configuration will also have unpaired electron irrespective of SFL and WFL. Co ^ 3+ [ Ar ] 3 ~d ^6 d ^6 it can be both paramagnetic or diamagnetic based on field of ligands. In case of F ^ - weak field ligand, configuration will be t _ 2 ~g ^4 e _ g ^2 hence it is paramagnetic but in case of en strong filed ligand, configuration wi

Q: The complex(es), which can exhibit the type of isomerism shown by Pt NH 3 2 Br 2 , is(are) en = H 2 NCH 2 CH 2 NH 2

Pt NH 3 2 Cl 2 is a square planar complex and it can exhibit geometrical isomerism. It has two isomers, cis-platin and trans-platin. The complexes Zn NH 3 2 Cl 2 tetrahedral and Pt ( en ) ( SCN ) 2 does not exhibit geometrical isomerism. The octahedral complex Pt NH 3 2 Cl 4 shows geometrical isomerism, it has two geometrical isomers cis and trans. Cr ( en ) 2 H 2 O SO 4 + is an octahedral complex, it is of the type M AA

Question 1

Reaction of PtF _6 with oxygen ( O _2) gas results in the formation of an ionic compound, X ^+ Y ^-. Correct statement(s) is(are)

Accepted Answer

The reaction between O _2 and PtF _6 is: O _2 + PtF _6 O _2^+[ PtF _6]^- Here, X^+ is O _2^+ and Y^- is [ PtF _6]^-. For O _2^+, the total number of electrons is 15. The molecular orbital configuration is _ 1s ^2 _ 1s ^ *2 _ 2s ^2 _ 2s ^ *2 _ 2p_z ^2 _ 2p_x ^2 _ 2p_y ^2 _ 2p_x ^ *1 . The bond order is 10 - 5 2 = 2.5. Thus, the bond order is not 1.5. In the complex ion [ PtF _6]^-, the oxidation state of Pt is +5. The ground state electronic configuration of Pt is [ Xe ] 4f^ 14 5d^9 6s^1. The con

Question 2

The total number of all possible isomers for the square planar complex with formula K [ M(NCS)(NO _2 )(gly) ] is ____. (M = metal ion and gly = NH _2 CH _2 COO ^-)

Accepted Answer

The given complex is K [ M(NCS)(NO _2 )(gly) ]. The complex anion is [ M(NCS)(NO _2 )(gly) ]^ - . The ligands present in the complex are: 1. gly ^- (glycinate): An unsymmetrical bidentate ligand coordinating through nitrogen (N) and oxygen (O). Let this be represented as (AB). 2. NCS ^-: An ambidentate monodentate ligand that can coordinate through nitrogen ( -NCS ) or sulfur ( -SCN ). 3. NO _2^-: An ambidentate monodentate ligand that can coordinate through nitrogen ( -NO _2) or oxygen ( -ONO )

Question 3

Consider that the coordinating atoms of the ligands in cis-[Co(NH_3)_4Cl_2]Cl and mer-[Co(NH_3)_3Cl_3] octahedral complexes are at the vertices of an octahedron. The sum of total number of the triangu

Accepted Answer

In an octahedral complex, the six coordinating atoms are located at the vertices of an octahedron, which has 8 triangular faces. Each face is formed by 3 vertices, and any two adjacent (cis) vertices form an edge that is shared by exactly two faces. For the cis-[Co(NH_3)_4Cl_2]^+ complex: The two Cl atoms are at adjacent (cis) positions, forming exactly one Cl-Cl edge. This edge is shared by two triangular faces. Since the remaining four vertices are occupied by N atoms, the third vertex of each

Question 4

The correct order of the wavelength maxima of the absorption band in the ultraviolet-visible region for the given complexes is

Accepted Answer

_ O 1 aligned & The absorb wave length order is \ & [ Co ( CN )_6 ]^ 3- < [ Co ( NH _3 )_6 ]^ 3+ < [ Co ( NH _3 )_5 H _2 O ]^ 3+ < [ Co ( NH _3 )_5 Cl ]^ 2+ aligned

Question 5

The sum of the spin only magnetic moment values (in B.M.) of [ Mn ( Br )_6 ]^ 3- and [ Mn ( CN )_6 ]^ 3- is _____.

Accepted Answer

Sum of spin magnetic moments of both complexes is 7.70 to 7.73 ~B . M .

Question 6

Among [ Co ( CN )_4 ]^ 4- , [ Co ( CO )_3( NO ) ], XeF _4, [ PCl _4 ]^ + , [ PdCl _4 ]^ 2- , [ ICl _4 ]^ - , [ Cu ( CN )_4 ]^ 3- and P _4 the total number of species with tetrahedral geometry is _____

Accepted Answer

[ Co ( CN )_4 ]^ 4- , [ Co ( CO )_3 NO ], [ PCl _4 ]^ + , [ Cu ( CN )_4 ]^ 3- ~\&~ P _4 are with tetrahedral geometry.

Question 7

Among the following options, select the option in which each complex in Set-I shows geometrical isomerism and the two complexes in Set-II are ionization isomers of each other. [en .= H _2 NCH _2 CH _2

Accepted Answer

Set-I: $ [ Co ( NH _3 )_3 ( NO _2 )_3 ]$ and $ [ Co (e n)_2 Cl _2 ]$ Set-II: $ [ Co ( NH _3 )_5 Cl ] SO _4$ and $ [ Co ( NH _3 )_5 ( SO _4 ) ] Cl $ Set-I: Set-II: $ [ Co ( NH _3 )_5 Cl ] SO _4$ and $ [ Co ( NH _3 )_5 SO _4 ] Cl $ are ionisation isomers.

Question 8

Among V ( CO )_6, Cr ( CO )_5, Cu ( CO )_3, Mn ( CO )_5, Fe ( CO )_5, [ Co ( CO )_3 ]^ 3- , [ Cr ( CO )_4 ]^ 4- , and Ir ( CO )_3, the total number of species isoelectronic with Ni ( CO )_4 is _______

Accepted Answer

In case of complexes, isoelectronic species should be those having same effective atomic number (EAN) Ni ( CO )_4 28+4 2=36 (i) V ( CO )_6 23+2 6=35 (ii) Cr ( CO )_5 24+2 5=34 (iii) Cu ( CO )_3 29+2 3=35 (iv) Mn ( CO )_5 25+2 5=35 (v) Fe ( CO )_5 26+2 5=36 (vi) [ Co ( CO )_3 ]^ 3- 27+3+2 3=36 (vii) [ Cr ( CO )_4 ]^ 4- 24+4+2 4=36 (viii) [ Ir ( CO )_3 ] 77+2 3=83

Question 9

Among the following complexes, the total number of diamagnetic species is_____ [ Mn ( NH _3 )_6 ]^ 3+ , [ MnCl _6 ]^ 3- , [ FeF _6 ]^ 3- , [ CoF _6 ]^ 3- , [ Fe ( NH _3 )_6 ]^ 3+ , and [ Co ( en )_3 ]

Accepted Answer

Mn ^ 3+ [ Ar ] 3 ~d ^4 d ^4 configuration in t _ 2 ~g and eg _ g orbitals will always have unpaired electrons irrespective of SFL and WFL. Fe ^ 3+ [ Ar ] 3 ~d ^5 d ^5 configuration will also have unpaired electron irrespective of SFL and WFL. Co ^ 3+ [ Ar ] 3 ~d ^6 d ^6 it can be both paramagnetic or diamagnetic based on field of ligands. In case of F ^ - weak field ligand, configuration will be t _ 2 ~g ^4 e _ g ^2 hence it is paramagnetic but in case of en strong filed ligand, configuration wi

Question 10

The complex(es), which can exhibit the type of isomerism shown by Pt NH 3 2 Br 2 , is(are) en = H 2 NCH 2 CH 2 NH 2

Accepted Answer

Pt NH 3 2 Cl 2  is a square planar complex and it can exhibit geometrical isomerism. It has two isomers, cis-platin and trans-platin. The complexes  Zn NH 3 2 Cl 2 &#160; tetrahedral &#160; and &#160; Pt ( en ) ( SCN ) 2  does not exhibit geometrical isomerism. The octahedral complex  Pt NH 3 2 Cl 4  shows geometrical isomerism, it has two geometrical isomers cis and trans.   Cr ( en ) 2 H 2 O SO 4 +  is an octahedral complex, it is of the type  M AA

Question 11

Match the electronic configurations in List-I with appropriate metal complex ions in List-II and choose the correct option. [Atomic Number: Fe &#160; = &#160; 26 , &#160; Mn &#160; = &#160; 25 , &#160

Accepted Answer

In the complex  Fe H 2 O 6 2 +  the electronic configuration of  Fe 2 +  is  Ar 3 d 6 .  H 2 O  is a weak ligand. Hence, the crystal filed splitting can be shown as follows, In the complex  Mn H 2 O 6 2 +  the electronic configuration of  Mn 2 +  is  Ar 3 d 5 .  H 2 O  is a weak ligand. Hence, the crystal filed splitting can be shown as follows, In the complex  Co NH 3 6 3 +  the electronic configuration of  Fe

Question 12

LIST- I contains metal species and LIST- II  contains their properties.   LIST- I   List- II I Cr CN 6 4 - P t 2 g  orbitals contain  4  electrons II RuCl 6 2 - Q &#956;

Accepted Answer

I &#160; &#160; Cr CN 6 4 - Cr 2 = Ar 3 d 4 4 s 0 It is  d 2 sp 3  hybridised low spin complex as  CN -  is a strong field ligand. II &#160; &#160; RuCl 6 2 - Low spin complex as Ru belongs to 4d series, as elements belonging to 4d and 5d series forms low spin complexes even with weak field ligand. Ru + 4 = Kr 4 d 4 5 s 0 t 2 g  set contains  4  electrons. III &#160; &#160; Cr H 2 O 6 2 + Cr + 2 = Ar 3 d 4 4 s 0 Its configuration is  t 2 g 3 e g 1 It

Question 13

The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are) (Note: py = pyridine Given: Atomic numbers of Fe , &#160; Co , &#160; Ni and Cu are 26 , &#160; 27 , &#160; 28 and 29 , respe

Accepted Answer

FeCl 4 - :  sp 3  Tetrahedral Fe CO 4 2 -  :  sp 3  Tetrahedral (Isoelectronic to  Ni CO 4 ) Co CO 4 -  :  sp 3  Tetrahedral (Isoelectronic to  Ni CO 4 ) CoCl 4 2 -  :  sp 3  Tetrahedral Ni Co 4  :  sp 3  Tetrahedral (Complex of Mond's process) Ni CN 4 2 -  :  dsp 2  square planar (Cyanide is strong field ligand) Cu Py 4 +  :  sp 3  Tetrahedral Cu CN 4 3 -  :  sp 3 &nb

Question 14

The calculated spin only magnetic moments of  Cr NH 3 6 3 +  and  CuF 6 3 -  in  BM ,  respectively, are (Atomic numbers of Cr and Cu are 24 and 29 , respectively)

Accepted Answer

Spin only magnetic moment can be calculated using the formula  &#956; = n &#160; n + 2   B . M n =  number of unpaired electron at central atom. Cr NH 3 6 3 +  :  Cr 3 +  is the central metal ion. Cr 3 +  :  3 d 3  configuration  It has three unpaired electrons. &#956; = 3 3 + 2 &#160; BM = 3 .87 &#160; BM CuF 6 3 &#8722;  :  Cu 3 +  is the central metal ion. Cu 3 +  :  3 d 8  configuration Fluorine is the weak

Question 15

The total number of possible isomers for  Pt NH 3 4 Cl 2 Br 2  is___.

Accepted Answer

Pt NH 3 4 Cl 2 Br 2  exhibits both ionisation and Geometrical isomerism : Pt NH 3 4 Cl 2 Br 2 ,  Pt NH 3 4 ClBr BrCl  and  Pt NH 3 4 Br 2 Cl 2  are the ionisation isomers and each isomer exhibits geometrical isomerism. Hence, total number of isomers  = 6 .

Question 16

Choose the correct statement(s) among the following:

Accepted Answer

( A ) FeCl 4 - Oxidation number of Fe atom = + 3 Fe → 3 d 6 4 s 2 Fe 3 + → 3 d 5 ∴ FeCl 4 - is having tetrahedral geometry ( B ) Co ( en ) NH 3 2 + have three geometrical isomers. ( C ) FeCl 4 - is sp 3 hybridized with 5 -unpaired electrons. While Co ( en ) NH 3 2 Cl 2 + is d 2 sp 3 hybridised with zero unpaired electrons. ( D ) Co ( en ) NH 3 2 Cl 2 + has d 2 sp 3 hybridization.

Question 17

Total number of cis N - M n - C l bond angles (that is, M n - N and M n - C l bonds in cis positions) present in a molecule of cis - M n e n 2 C l 2 complex is _________ e n = N H 2 C H 2 C H 2 N H 2

Accepted Answer

Total N - M n - C l bonds = 6.00

Coordination Compounds — Previous Year Questions & Solutions