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301. In an anionic coordination entity containing lead, the metal name becomes:
ⓐ. plumbate
ⓑ. leadate
ⓒ. plumbide
ⓓ. plumbo
Correct Answer: plumbate
Explanation: Lead takes the special name plumbate in anionic coordination entities. This is one of the metal-name exceptions that must be remembered in systematic coordination nomenclature. It follows the same pattern as ferrate, cuprate, argentate, aurate, and stannate. So the correct anionic coordination name for lead uses plumbate.
302. Which square planar complex does not show geometrical isomerism because all four coordinated ligands are identical?
ⓐ. \([Pt(NH_3)_2Cl_2]\)
ⓑ. \([Pt(NH_3)_3Cl]^+\)
ⓒ. \([PtCl_4]^{2-}\)
ⓓ. \([Pt(NH_3)_2BrCl]\)
Correct Answer: \([PtCl_4]^{2-}\)
Explanation: A square planar complex with four identical ligands has only one possible arrangement around the metal. Since there is no difference between adjacent and opposite placement of identical groups, geometrical isomerism does not arise. In contrast, complexes such as \([Pt(NH_3)_2Cl_2]\) can show cis and trans forms because they contain two different ligand types. So \([PtCl_4]^{2-}\) is non-isomeric in this sense.
303. Which set of special metal names for anionic coordination entities is correctly matched?
ⓐ. iron \(\rightarrow\) ironate, copper \(\rightarrow\) copperate, silver \(\rightarrow\) silverate
ⓑ. iron \(\rightarrow\) ferrate, copper \(\rightarrow\) cuprate, silver \(\rightarrow\) argentate
ⓒ. iron \(\rightarrow\) ferride, copper \(\rightarrow\) cupride, silver \(\rightarrow\) argentide
ⓓ. iron \(\rightarrow\) ferrito, copper \(\rightarrow\) cuprito, silver \(\rightarrow\) argentito
Correct Answer: iron \(\rightarrow\) ferrate, copper \(\rightarrow\) cuprate, silver \(\rightarrow\) argentate
Explanation: Some metals take special traditional names when they occur in anionic coordination entities. Iron becomes ferrate, copper becomes cuprate, and silver becomes argentate. These forms are standard in coordination nomenclature and must be remembered directly. They are not obtained by simply adding \(-ate\) to the ordinary English metal name in the usual way.
304. Which square planar complex does not show geometrical isomerism because its ligand arrangement allows only one distinct form?
ⓐ. \([Pt(NH_3)_2Cl_2]\)
ⓑ. \([Pt(NH_3)_3Cl]^+\)
ⓒ. \([Pt(NH_3)_2BrCl]\)
ⓓ. \([Pt(NH_3)_2Br_2]\)
Correct Answer: \([Pt(NH_3)_3Cl]^+\)
Explanation: A square planar complex of the type \([MA_3B]\) gives only one arrangement. There is no second distinct adjacent-versus-opposite pattern to compare, so cis-trans isomerism does not arise. In contrast, square planar complexes such as \([Pt(NH_3)_2Cl_2]\), \([Pt(NH_3)_2BrCl]\), and \([Pt(NH_3)_2Br_2]\) have arrangements that can give distinguishable geometrical forms. The key reason is the presence of two pairs or two different ligand types that allow alternative spatial placements.
305. Which statement is correct for a tetrahedral complex of the type \([MA_2B_2]\) with simple monodentate ligands?
ⓐ. It usually shows fac-mer isomerism
ⓑ. It usually shows only optical isomerism
ⓒ. It always gives three geometrical isomers
ⓓ. It usually lacks geometrical isomerism
Correct Answer: It usually lacks geometrical isomerism
Explanation: In tetrahedral geometry, there is no clear adjacent-versus-opposite distinction like the one seen in square planar or octahedral complexes. Because of that, a tetrahedral \([MA_2B_2]\) arrangement with simple monodentate ligands does not normally produce distinct cis and trans forms. This is one of the standard restrictions on geometrical isomerism. It is an important contrast with square planar \([MA_2B_2]\) complexes.
306. Which tetrahedral coordination arrangement can show optical isomerism?
ⓐ. A tetrahedral complex with four different ligands around the metal
ⓑ. A tetrahedral complex with two pairs of identical ligands
ⓒ. A tetrahedral complex with four identical ligands
ⓓ. A tetrahedral complex containing only one ligand outside the bracket
Correct Answer: A tetrahedral complex with four different ligands around the metal
Explanation: A tetrahedral complex can become chiral when all four groups around the metal are different. In that case, the complex and its mirror image cannot be superimposed on one another. This gives optical isomerism. When ligands are repeated in a symmetrical way, that chirality is usually lost.
307. Optical activity in a coordination compound generally requires the absence of:
ⓐ. a metal ion
ⓑ. a coordination sphere
ⓒ. a plane of symmetry
ⓓ. any covalent character in the metal-ligand bond
Correct Answer: a plane of symmetry
Explanation: A coordination compound is optically active when it is chiral, meaning its mirror image cannot be placed exactly on top of it. A plane of symmetry often destroys chirality because it makes the structure equivalent to its mirror image. So the absence of such symmetry is an important condition for optical activity. This is why many highly symmetrical complexes are optically inactive.
308. Which complex is expected to show optical isomerism?
ⓐ. trans-\([Co(en)_2Cl_2]^+\)
ⓑ. cis-\([Co(en)_2Cl_2]^+\)
ⓒ. \([Co(NH_3)_6]^{3+}\)
ⓓ. trans-\([Pt(NH_3)_2Cl_2]\)
Correct Answer: cis-\([Co(en)_2Cl_2]^+\)
Explanation: The cis form of \([Co(en)_2Cl_2]^+\) can have a chiral arrangement because the two bidentate \(en\) ligands wrap around the metal in a way that gives non-superimposable mirror images. The trans form is usually not taken as optically active at this level because of its greater symmetry. The highly symmetrical \([Co(NH_3)_6]^{3+}\) also does not show optical isomerism here. So the cis chelate complex is the appropriate example.
309. For successive formation of \(ML\), \(ML_2\), and \(ML_3\) from the same metal ion and ligand, which trend is generally expected for the stepwise stability constants?
ⓐ. \(K_1 = K_2 = K_3\)
ⓑ. \(K_3 > K_2 > K_1\)
ⓒ. \(K_2 > K_1 > K_3\)
ⓓ. \(K_1 > K_2 > K_3\)
Correct Answer: \(K_1 > K_2 > K_3\)
Explanation: In many successive complex-formation steps, the first ligand binds most easily because all coordination positions are initially available. As ligands are added, fewer free positions remain and further addition often becomes less favourable. So the stepwise stability constants commonly decrease in order. This broad trend is written as \(K_1 > K_2 > K_3\).
310. For comparable ligands and oxidation states, which statement is generally correct about \(3d\), \(4d\), and \(5d\) metal ions?
ⓐ. \(4d\) and \(5d\) ions more often form low-spin complexes than comparable \(3d\) ions
ⓑ. \(3d\) metal ions always form low-spin complexes, while \(4d\) and \(5d\) ions always form high-spin complexes
ⓒ. \(4d\) and \(5d\) metal ions never use \(d\)-orbitals in bonding
ⓓ. All three series always show the same spin behaviour with every ligand
Correct Answer: \(4d\) and \(5d\) ions more often form low-spin complexes than comparable \(3d\) ions
Explanation: A broad trend is that \(4d\) and \(5d\) metals often experience stronger splitting effects than comparable \(3d\) metals. Because of that, electron pairing can become more favourable in many related cases. This makes low-spin behaviour more common for \(4d\) and \(5d\) series. The trend is a general comparison rather than an absolute rule for every complex.
311. Which limitation of valence bond theory is correctly stated?
ⓐ. It cannot identify the geometry of any coordination compound
ⓑ. It cannot distinguish a ligand from a counter ion
ⓒ. It does not explain ligand field strength satisfactorily
ⓓ. It cannot describe the existence of coordinate bonds
Correct Answer: It does not explain ligand field strength satisfactorily
Explanation: Valence bond theory gives a simple picture of bonding and geometry, but it does not satisfactorily explain the relative field strengths of ligands. It does not clearly show why ligands such as \(CN^-\) and \(CO\) produce much larger splitting than ligands such as \(F^-\) or \(Cl^-\). Crystal field theory improves on this point by relating ligand strength to \(d\)-orbital splitting. That is why VBT is considered limited in this respect.
312. In the usual spectrochemical series, which ligand is weaker than \(NH_3\) but stronger than \(F^-\)?
ⓐ. \(CN^-\)
ⓑ. \(CO\)
ⓒ. \(H_2O\)
ⓓ. \(I^-\)
Correct Answer: \(H_2O\)
Explanation: In the common spectrochemical series, fluoride lies on the weaker-field side, while ammonia lies further toward the stronger-field side. Water is placed between \(F^-\) and \(NH_3\). This means \(H_2O\) produces a larger splitting than \(F^-\), but a smaller splitting than \(NH_3\), for comparable metal ions and geometry. Cyanide and carbon monoxide are stronger-field ligands than ammonia, while iodide is weaker than fluoride. Therefore \(H_2O\) is the ligand that fits the stated comparison.
313. In an anionic coordination entity containing silver, the metal name becomes:
ⓐ. silverate
ⓑ. argentate
ⓒ. argentide
ⓓ. silvate
Correct Answer: argentate
Explanation: Some metals use special traditional names when they appear in anionic coordination entities. Silver is named argentate in such complexes. This is the same naming pattern seen in examples such as ferrate for iron and cuprate for copper. The form silverate is not the preferred systematic form at this level. Therefore an anionic silver complex uses the metal name argentate.
314. What is the correct name of \(K[Au(CN)_2]\)?
ⓐ. potassium dicyanidoaurate(I)
ⓑ. potassium dicyanidogold(III)
ⓒ. potassium cyanidoaurate(II)
ⓓ. dicyanidopotassiumaurate(I)
Correct Answer: potassium dicyanidoaurate(I)
Explanation: \(\textbf{Given:}\)
Compound \(K[Au(CN)_2]\)
\(\textbf{Required:}\)
Correct coordination name
\(\textbf{Identify the coordination entity:}\)
The complex ion is \([Au(CN)_2]^-\), and potassium is the counter ion.
\(\textbf{Find the oxidation state of gold:}\)
Let the oxidation state of gold be \(x\).
\[x + 2(-1) = -1\]
\[x = +1\]
\(\textbf{Apply naming rules:}\)
The anionic complex uses aurate for gold, and two \(CN^-\) ligands are named dicyanido.
\(\textbf{Final Answer:}\)
The compound is potassium dicyanidoaurate(I).
315. How many geometrical isomers are possible for a square planar complex of the type \([MA_2B_2]\)?
ⓐ. \(1\)
ⓑ. \(2\)
ⓒ. \(3\)
ⓓ. \(4\)
Correct Answer: \(2\)
Explanation: In a square planar complex of the type \([MA_2B_2]\), the two identical \(A\) ligands and the two identical \(B\) ligands can be arranged in two distinct ways. If the identical ligands are adjacent, the form is called cis. If they are opposite to each other, the form is called trans. These two arrangements have the same connectivity but different spatial arrangement. Therefore a square planar \([MA_2B_2]\) complex can show two geometrical isomers.
316. The stepwise stability constants for a complex are \(K_1 = 1.0 \times 10^4\), \(K_2 = 2.0 \times 10^3\), and \(K_3 = 5.0 \times 10^2\). What is the overall stability constant \(\beta_3\)?
ⓐ. \(1.0 \times 10^7\)
ⓑ. \(1.0 \times 10^9\)
ⓒ. \(1.0 \times 10^{10}\)
ⓓ. \(1.0 \times 10^{13}\)
Correct Answer: \(1.0 \times 10^{10}\)
Explanation: \(\textbf{Given:}\)
\(K_1 = 1.0 \times 10^4\)
\(K_2 = 2.0 \times 10^3\)
\(K_3 = 5.0 \times 10^2\)
\(\textbf{Required:}\)
Overall stability constant, \(\beta_3\)
\(\textbf{Relevant Formula:}\)
\[
\beta_3 = K_1K_2K_3
\]
\(\textbf{Why this formula applies:}\)
The overall stability constant is obtained by multiplying all successive stepwise constants for formation of the final complex.
\(\textbf{Substitution:}\)
\[
\beta_3 = (1.0 \times 10^4)(2.0 \times 10^3)(5.0 \times 10^2)
\]
\(\textbf{Intermediate Simplification:}\)
\[
(1.0 \times 2.0 \times 5.0) \times 10^{4+3+2}
\]
\[
10.0 \times 10^9
\]
\(\textbf{Final Simplification:}\)
\[
1.0 \times 10^{10}
\]
\(\textbf{Final Answer:}\)
\[
\beta_3 = 1.0 \times 10^{10}
\]
317. For a complex, \(\beta_2 = 6.0 \times 10^8\) and \(K_1 = 3.0 \times 10^4\). What is the value of \(K_2\)?
ⓐ. \(2.0 \times 10^4\)
ⓑ. \(2.0 \times 10^3\)
ⓒ. \(5.0 \times 10^4\)
ⓓ. \(1.8 \times 10^{13}\)
Correct Answer: \(2.0 \times 10^4\)
Explanation: \(\textbf{Given:}\)
\(\beta_2 = 6.0 \times 10^8\)
\(K_1 = 3.0 \times 10^4\)
\(\textbf{Required:}\)
Second stepwise stability constant, \(K_2\)
\(\textbf{Relevant Formula:}\)
\[
\beta_2 = K_1K_2
\]
\(\textbf{Why this formula applies:}\)
For formation up to \(ML_2\), the overall constant is the product of the first two stepwise constants.
\(\textbf{Rearrangement:}\)
\[
K_2 = \frac{\beta_2}{K_1}
\]
\(\textbf{Substitution:}\)
\[
K_2 = \frac{6.0 \times 10^8}{3.0 \times 10^4}
\]
\(\textbf{Intermediate Simplification:}\)
\[
K_2 = \left(\frac{6.0}{3.0}\right)\times 10^{8-4}
\]
\[
K_2 = 2.0 \times 10^4
\]
\(\textbf{Final Answer:}\)
\[
K_2 = 2.0 \times 10^4
\]
318. A ligand that can bind to the same metal ion through three donor atoms is called:
ⓐ. monodentate
ⓑ. bidentate
ⓒ. hexadentate
ⓓ. tridentate
Correct Answer: tridentate
Explanation: Denticity refers to the number of donor atoms of a ligand that attach to the same metal centre. A monodentate ligand uses one donor atom, a bidentate ligand uses two, and a tridentate ligand uses three. This classification is important because coordination number depends on donor atoms, not just on how many ligands are present. So a three-point attachment ligand is called tridentate.
319. A coordination entity contains one tridentate ligand, one bidentate ligand, and one monodentate ligand. What is its coordination number?
ⓐ. \(3\)
ⓑ. \(6\)
ⓒ. \(5\)
ⓓ. \(4\)
Correct Answer: \(6\)
Explanation: \(\textbf{Given:}\)
One tridentate ligand
One bidentate ligand
One monodentate ligand
\(\textbf{Required:}\)
Coordination number of the metal
\(\textbf{Relevant Principle:}\)
Coordination number is the total number of donor atoms directly attached to the metal ion.
\(\textbf{Identify donor contributions:}\)
Tridentate ligand contributes \(3\)
Bidentate ligand contributes \(2\)
Monodentate ligand contributes \(1\)
\(\textbf{Substitution:}\)
\[
\text{Coordination number} = 3+2+1
\]
\(\textbf{Final Simplification:}\)
\[
\text{Coordination number} = 6
\]
\(\textbf{Why }3\textbf{ is a common error:}\)
That counts only the number of ligands and ignores denticity.
\(\textbf{Final Answer:}\)
Coordination number \(= 6\)
320. In complexometric titrations, \(EDTA^{4-}\) is especially useful because it:
ⓐ. acts only as a weak-field ligand
ⓑ. forms unstable complexes that dissociate rapidly
ⓒ. forms stable chelates with many metal ions
ⓓ. can coordinate only with alkali metals
Correct Answer: forms stable chelates with many metal ions
Explanation: \(EDTA^{4-}\) is a hexadentate ligand and can bind to a metal ion through several donor atoms at once. This leads to formation of highly stable chelate complexes. Because of that strong and predictable binding, it is widely used in complexometric analysis. Its importance in estimation of metal ions and water hardness comes from this stability.