Download App

Coordination Compounds — Chemistry STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 ScienceChemistryCoordination Compounds4 Marks
Question
Read the passage given below and answer the following questions:
To explain bonding in coordination compounds various theories were proposed. One of the important theory was valence bond theory. According to that, the central metal ion in the complex makes available a number of empty orbitals for the formation of coordination bonds with suitable ligands. The appropriate atomic orbitals of the metal hybridise to give a set of equivalent orbitals of definite geometry.
The d-orbitals involved in the hybridisation may be either inner d-orbitals i.e., (n - 1)d or outer d-orbitals i.e., nd. For example, Co3+ forms both inner orbital and outer orbital complexes, with ammonia it forms [Co(NH3)6]3+ and with fluorine it forms [CoF6]3- complex ion.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following is not true for [CoF6]3-?
  1. It is paramagnetic.
  2. It has coordination number of 6.
  3. It is outer orbital complex.
  4. It involves d2sp3 hybridisation.
  1. [Cr(H2O)6]Cl3 (at. no. of Cr = 24) has a magnetic moment of 3.83B.M. The correct distribution of 3d-electrons in the central metal of the complex is:
  1. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{x}^2-\text{y}^2},3\text{d}^1_\text{yz}$
  2. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{yz}},3\text{d}^1_\text{zx}$
  3. $3\text{d}^1_\text{xy},3\text{d}^1_{\text{zy}},3\text{d}^1_{\text{z}^2}$
  4. $3\text{d}^1_{\text{x}^2-\text{y}^2},3\text{d}^1_{\text{z}^2},3\text{d}^1_\text{xz}$
  1. Which of the following is true for [Co(NH3)6]3+?
  1. It is an octahedral, di magnetic and outer orbital complex.
  2. It is an octahedral, paramagnetic and outer orbital complex.
  3. It is an octahedral, paramagnetic and inner orbital complex.
  4. It is an octahedral, di magnetic and inner orbital complex.
  1. The paramagnetism of [CoF6]3- is due to.
  1. 3 electrons.
  2. 4 electrons.
  3. 2 electrons.
  4. 1 electron.
  1. Which of the following is an inner orbital or low spin complex?
  1. [Ni(H2O)6]3+
  2. [FeF6]3-
  3. [Co(CN)6]3-
  4. [NiCl4]2-
✓

Answer

  1. (d) It involves d2sp3 hybridisation.

Explanation:

lt involves sp3d2 hybridisation and not d2sp3.

  1. (b) $3\text{d}^1_\text{xy},3\text{d}^1_{\text{yz}},3\text{d}^1_\text{zx}$

Explanation:

Magnetic moment of 3.83B.M. suggests that it has 3 unpaired electrons,

$\therefore$ n = 3 i.e., Cr3+: 3d3

It involves d2sp3 hybridisation so correct distribution of electrons is $3\text{d}^1_\text{xy},3\text{d}^1_{\text{yz}},3\text{d}^1_\text{zx}.$

  1. (d) It is an octahedral, di magnetic and inner orbital complex.

Explanation:

[Co(NH3)6]3+ is d2sp3 hybridised with all electrons paired hence, it is diamagnetic and inner orbital complex.

  1. (b) 4 electrons.

Explanation:

  1. (c) [Co(CN)6]3-

Explanation:

Inner orbital complexes are formed with strong ligands as they force electrons to pair up and hence the complex will be either diamagnetic or will have less number of unpaired electrons.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "Case study (4 Marks)" from Coordination Compounds→Coordination Compounds — all question types→Chemistry STD 12 Science question papers→Bihar Board STD 12 Science question papers→Bihar Board question paper generator→

Similar questions

Read the passage given below and answer the following questions:

The order of reactivity towards SN1 reaction depends upon the stability of carbocation in the first step. Greater the stability of the carbocation, greater will be its ease of formation from alkyl halide and hence faster will be the rate of the reaction. As we know, 3º carbocation is most stable, therefore, the tert-alkyl that halides will undergo SN1 reaction very fast. For example, it has been observed that the reaction (CH3)3CBr with OH- ion to give 2-methyl-2-propanol is about I million times as fast as the corresponding reaction of the methyl bromide to give methanol.

The primary alkyl halides always react predominantly by SN2 mechanism. On the other hand, the tertiary alkyl halides react predominantly by SN1 mechanism. Secondary alkyl halides may react by either mechanism or by both the mechanisms without much preference depending upon the nature of the nucleophile and solvent.

In these questions (Q. No. i-tv), a statement of assertion followed by a statement of reason is given. Choose tile correct answer out of tile following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Low concentration of nudeophile favours SN1 mechanism.

Reason: 2º alkyl halides are less reactive than 1º towards SN1 reactions.

  1. Assertion: Polar solvent slows down SN2 reactions.

Reason: CH3-Br is less reactive than CH3Cl.

  1. Assertion: Benzyl bromide when kept in acetone- water it produces benzyl alcohol.

Reason: The reaction follows SN2 mechanism.

  1. Assertion: Rate of hydrolysis of methyl chloride to methanol is higher in DMF than in water.

Reason: Hydrolysis of methyl chloride follows second order kinetics.

  1. Assertion: SN1 reaction is carried out in the presence of a polar protic solvent.

Reason: A polar protic solvent increases the stability of carbocation due to solvation.

→

Read the passage given below and answer the following questions:

Amines are basic in nature. The basic strength of amines can be expressed by their dissociation constant, Kb or pKb.

$\text{RNH}_2+\text{H}_2\text{O}\rightleftharpoons\text{RNH}^+_3+\text{OH}^-$

$\text{k}_\text{b}=\frac{[\text{RNH}^+_3][\text{OH}^-]}{[\text{RNH}_2]}\text{and}\text{ pk}_\text{b}=-\log\text{k}_\text{b}$

Greater the Kb value or smaller the pKb value, more is the basic strength of a mine. Aryl amines such as aniline are less basic than aliphatic amines due to the involvement of lone pair of electrons on N-atom with the resonance in benzene. In derivatives of aniline, the electron releasing groups increase the basic strength while electron withdrawing groups decrease the basic strength. The base weakening effect of electron withdrawing group and base strengthening effect of electron releasing group is more marked at p-position than at m-position. a-Substituted aniline is less basic than aniline due to ortho effect and is probable due to combination of electronic and steric effect.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Which of the following has lowest pKb value?

  1. The strongest base among the following is:

  1. C6H5NH2

  2. p-NO2 - C6H4NH2

  3. m-NO2 - C6H4NH2

  4. C6H5CH2NH2

  1. Maximum pKb value of:
  1.  

  1.  

  1. (CH3CH2)2NH

  2. (CH3)2NH

  1.  The order of basic strength among the following amines in benzene solution is:
  1. Methylamine is more basic than NH3.
  2. Amines form hydrogen bonds.
  3. Ethylamine has higher boiling point than propane.
  4. Dimethylamine is less basic than methylamine.
  1. CH3CH2NH2 contains a basic -NH2 group, but CH3CONH2 does not because:
  1. Acetamide is amphoteric in character.
  2. In ethylamine the electron pair on N-atom is delocalised by resonance.
  3. In ethylamine there is no resonance while in acetamide the lone pair of electrons on N-atom is delocalised and is less available for protonation.
  4. None of these.
→

Read the passage given below and answer the following questions:

Both alcohols and phenols are acidic in nature, but phenols are more acidic than alcohols. Acidic strength of alcohols mainly depends upon the inductive effect. Acidic strength of phenols depends upon a combination of both inductive effect and resonance effects of the substituent and its position on the benzene ring. Electron withdrawing groups increases the acidic strength of phenols whereas electron donating groups decreases the acidic strength of phenols. Phenol is a weaker acid than carboxylic acid.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Phenols are highly acidic as compare to alcohols due to:
  1. The higher molecular mass of phenols.
  2. The stronger hydrogen bonds in phenols.
  3. Alkoxide ion is a strong conjugate base.
  4. Phenoxide ion is resonance stabilised.
  1. The correct order of acidic strength among the following is:

  1. (III) > (IV) > (II) > (I)
  2. (IV) > (III) > (I) > (II)
  3. (IV) > (III) > (II) > (I)
  4. (I) > (II) > (IV) > (III)
  1. The correct decreasing order of pKa value is:

  1. II > IV > I > III
  2. IV > II > III > I
  3. II I> II > IV > I
  4. IV > I > II > III
  1. The compound that does not liberate CO2, on treatment with aqueous sodium bicarbonate solution is:
  1. Benzoic acid.
  2. Benzenesulphonic acid.
  3. Salicylic acid.
  4. Carbolic acid.
  1. Most acidic amongst the following is:

→
Read the passage given below and answer the following questions:

Transition elements are elements that have partially filled d-orbitals. The configuration of these elements corresponds to (n - 1)d1-10ns1-2. It is important to note that the elements mercury, cadmium and zinc are not considered transition elements because of their electronic configurations, which corresponds to (n - 1)d10ns2. Some general properties of transition elements are:

These elements can fonn coloured compounds and ions due to d-d transition;

These elements exhibit many oxidation states;

A large variety of ligands can bind themselves to these elements, due to this, a wide variety of stable complexes formed by these ions. The boiling and melting point of these elements are high. These elements have a large ratio of charge to the radius.

In these questions (Q. No. i-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Tungsten has very high melting point.

Reason: Tungsten is a covalent compound.

  1. Assertion: Zn, Cd and Hg are normally not considered transition metals.

Reason: d-Orbitals in Zn, Cd and Hg elements are completely filled, hence these metals do not show the general characteristics properties of the transition elements.

  1. Assertion: Copper metal gets readily corroded in acidic aqueous solution such as HCl and dil. H2SO4.

Reason: Free energy change for this process is positive.

  1. Assertion: Tailing of mercury occurs on passing ozone through it.

Reason: Due to oxidation of mercury.

  1. Assertion: Transition metals are poor reducing agents.

Reason: Transition metals fonn numerous alloys with other metals.

→
Describe detailed information on classification of carbohydrates.
→
Explain the structure of nucleic acid compounds.
→
The progress of the reaction, $\text{A}\rightleftharpoons\text{nB}$ with time is represented in the following figure:

The following questions are multiple choice questions. Choose the most appropriate answer:
  1. What is the value of n?
  1. 1
  2. 2
  3. 3
  4. 4
  1. Find the value of the equilibrium constant.
  1. 0.6M
  2. 1.2M
  3. 0.3M
  4. 2.4M
  1. The initial rate of conversion of A will be:
  1. 0.1 mol L-1hr-1
  2. 0.2 mol L-1hr-1
  3. 0.4 mol L-1hr-1
  4. 0.8 mol L-1hr-1
  1. For the reaction, if $\frac{\text{d}[\text{B}]}{\text{dt}}=2\times10^{-4},$ value of $-\frac{\text{d}[\text{A}]}{\text{dt}}$ will be:
  1. 2 × 10-4
  2. 10-4
  3. 4 × 10-4
  4. 0.5 × 10-4
  1. Which factor has no effect on rate of reaction?
  1. Temperature.
  2. Nature of reactant.
  3. Concentration of reactant.
  4. Molecularity.
→
Read the passage given below and answer the following questions:
The unique behaviour of Cu, having a positive E° accounts for its inability to liberate H2 from acids. Only oxidising acids (nitric and hot concentrated sulphuric acid) react with Cu, the acids being reduced. The stability of the half-filled (d5) subshell in Mn2+ and the completely filled (d10) configuration in Zn2+ are related to their $\text{E}^\circ\frac{\text{M}^{3+}}{\text{M}^{2+}}$ values. The low value for Sc reflects the stability of Sc3+ which has a noble gas configuration. The comparatively high value for Mn shows that Mn2+(d5) is particularly stable, whereas a comparatively low value for Fe shows the extra stability of Fe3+(d5). The comparatively low value for Vis related to the stability of v2+ (half-filled t2g level).
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Standard reduction electrode potential of $\frac{\text{Zn}^{2+}}{\text{Zn}}$ is 0.76V. This means
  1. ZnO cannot be reduced to Zn by H2 under standard conditions.
  2. Zn cannot liberate H2 with concentrated acids.
  3. Zn is generally the anode in an electrochemical cell.
  4. Zn is generally the cathode in an electrochemical cell.
  1. $\text{E}^\circ$ values for the couples $\frac{\text{Cr}^{3+}}{\text{Cr}^{2+}}$ and $\frac{\text{Mn}^{3+}}{\text{Mn}^{2+}}$ are -0.41 and +1.51 volts respectively. These values suggest that.
  1. Cr2+ acts as a reducing agent whereas Mn3+ acts as an oxidizing agent.
  2. Cr2+ is more stable th an Cr3+ state.
  3. Mn3+ is more stable than Mn2+.
  4. Cr2+ acts as an oxidizing agent whereas Mn3+ acts as a reducing agent..
  1. The reduction potential values of M, N and O are +2.46, -1.13 and -3.13V respectively. Which of the following order is correct regarding their reducing property?
  1. O > N > M
  2. O > M > N
  3. M > N > O
  4. M > O > N
  1. Which of the following statements are true?
  1. Mn2+ compounds are more stable than Fe2+ towards oxidation to +3 state.
  2. Titanium and copper both in the first series of transition metals exhibits +1 oxidation state most frequently.
  3. Cu+ ion is stable in aqueous solutions.
  4. The $\text{E}^\circ$ value for the $\frac{\text{Mn}^{3+}}{\text{Mn}^{2+}}$ couple is much more positive than that for $\frac{\text{Cr}^{3+}}{\text{Cr}^{2+}}$ or $\frac{\text{Fe}^{3+}}{\text{Fe}^{2+}}.$.
  1. (II) and (III)
  2. (I) and (IV)
  3. (I) and (III)
  4. (II) and (IV)
  1. The stability of $\text{Cu}^{2+}_\text{(aq)}$ rather than $\text{Cu}^{+}_\text{(aq)}$ is due to.
  1. More negative $\Delta_\text{hyd}\text{H}^\circ$ of $\text{Cu}^{2+}_\text{(aq)}.$
  2. Less negative $\Delta_\text{hyd}\text{H}^\circ$ of $\text{Cu}^{2+}_\text{(aq)}.$
  3. More positive $\Delta_\text{hyd}\text{H}^\circ$ of $\text{Cu}^{2+}_\text{(aq)}.$
  4. Less positive $\Delta_\text{hyd}\text{H}^\circ$ of $\text{Cu}^{2+}_\text{(aq)}.$
→
Read the passage given below and answer the following questions:

Werner, a Swiss chemist in 1892 prepared and characterised a large number of coordination compounds and studied their physical and chemical behaviour. He proposed that, in coordination compounds, metals possess two types of valencies, viz. primary valencies, which are normally ionisable and secondary valencies which are non-ionisable. ln a series of compounds of cobalt (III) chloride with ammonia, it was found that some of the chloride ions could be precipitated as AgCl on adding excess of AgNO3 solution in cold, but some remained in solution. The number ofions furnished by a complex in a solution can be determined by precipitation reactions. The measurement of molar conductance of solutions of coordination compounds helps to estimate the number of ions furnished by the compound in solution.

In these questions (Q. No. i-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Assertion: The complex [Co(NH3)3Cl3] does not give precipitate with silver nitrate solution.

Reason: The given complex is non-ionisable.

  1. Assertion: The complex [Co(NH3)4Cl2]Cl gives precipitate corresponding to 2 mol of AgCl with AgNO3 solution.

Reason: It ionises as [Co(NH3)4Cl2]+ + Cl-.

  1. Assertion: CoCl3. 4NH3 gives 1 mol of AgCl on reacting with AgNO3, its secondary valency is 6.

Reason: Secondary valency corresponds to coordination number.

  1. Assertion: 1 mol of [CrCl2(H2O)4]Cl· 2H2O will give 1 mol of AgCl on treating with AgNO3.

Reason: Cl- ions satisfying secondary valanceis will not be precipitated.

  1. Assertion: CoCl3. 3NH3 is not conducting while CoCl3. 5NH3 is conducting.

Reason: The complex of CoCl3. 3NH3 is [CoCl3(NH3)3] while that of CoCl3· 5NH3 is [CoCl(NH3)5]Cl3.

→

Read the passage given below and answer the following questions:

Pentose and hexose undergo intramolecular hemiacetal or hemiketal formation due to combination of the –OH group with the carbonyl group. The actual structure is either of five or six membered ring containing an oxygen atom. In the free state all pentoses and hexoses exist in pyranose form (resembling pyran). However,inthe combined state some of them exist as five membered cyclic structures, called furanose (resembling furan).

The cyclic structure of glucose is represented by Haworth structure:

$\alpha$ and $\beta$ D-glucose have different configuration at anomeric (C-1) carbon atom, hence are called anomers and the C-1 carbon atom is called anomeric carbon (glycosidic carbon).

The six membered cyclic structure of glucose is called pyranose structure.

The following questionsare multiple choice questions. Choose the most appropriate answer:

  1. $\alpha$ D(+)-glucose and $\beta$ D(+)glucose are:
  1. Enantiomers.
  2. Conformers.
  3. Epimers.
  4. Anomers.
  1. The following carbohydrate is:

 

  1. A ketohexose.
  2. An aldohexose.
  3. An n-furanose.
  4. An $\alpha$-pyranose.
  1. In the following structure, anomeric carbon is:

  1. C-1
  2. C-2
  3. C-3
  4. C-4
  1. The term anomers of glucose refers to:
  1. Isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4).
  2. A mixture of (D)-glucose and (L)-glucose.
  3. Enantiomers of glucose.
  4. Isomers of glucose that differ in configuration at carbon one (C-1).
  1. What percentage of $\beta$-D-(+) glucopyranose is found at equilibrium in the aqueous solution?
  1. 50%

  2. $\approx100%$

  3. 36%
  4. 64%
→