301. Suitable \(d^8\) metal ions commonly form square-planar complexes with strong-field ligands because the ligands can:
ⓐ. pair electrons to free one inner \(d\) orbital
ⓑ. force the coordination number to become \(6\)
ⓒ. convert every metal \(d\) electron into a ligand electron
ⓓ. prevent the \(s\) and \(p\) orbitals from participating
Correct Answer: pair electrons to free one inner \(d\) orbital
Explanation: A \(d^8\) ion initially distributes eight electrons among five \(d\) orbitals. For \(\mathrm{dsp^2}\) hybridisation, one suitable inner \(d\) orbital must be vacant. Strong-field ligands can promote electron pairing in the remaining \(d\) orbitals. The vacant \(d\) orbital then mixes with one \(s\) and two \(p\) orbitals. This produces a square-planar arrangement that is commonly diamagnetic in standard \(d^8\) examples.
302. Which set of properties is most consistent with a standard strong-field square-planar \(d^8\) complex?
ⓐ. \(\mathrm{sp^3}\), tetrahedral, and two unpaired electrons
ⓑ. \(\mathrm{sp^3d^2}\), octahedral, and four unpaired electrons
ⓒ. \(\mathrm{dsp^2}\), square planar, and diamagnetic
ⓓ. \(\mathrm{d^2sp^3}\), octahedral, and diamagnetic
Correct Answer: \(\mathrm{dsp^2}\), square planar, and diamagnetic
Explanation: Strong-field ligands can cause the \(d^8\) electrons to pair in four inner \(d\) orbitals. The remaining vacant inner \(d\) orbital participates in \(\mathrm{dsp^2}\) hybridisation. Four coplanar hybrid orbitals are produced, giving square-planar geometry. Since all eight \(d\) electrons are paired, the standard complex is diamagnetic. The properties in option C are therefore mutually consistent.
303. A \(d^8\) metal ion forms a square-planar complex after ligand-induced pairing. Which orbital arrangement permits \(\mathrm{dsp^2}\) hybridisation?
ⓐ. All five inner \(d\) orbitals remain singly occupied
ⓑ. Four inner \(d\) orbitals are paired and one remains vacant
ⓒ. Two inner \(d\) orbitals remain vacant, while the remaining three contain eight electrons
ⓓ. All eight electrons move into the \(s\) and \(p\) orbitals
Correct Answer: Four inner \(d\) orbitals are paired and one remains vacant
Explanation: The metal ion has:
\[
d^8
\]
For square-planar bonding, one suitable inner \(d\) orbital must participate in hybridisation.
Strong ligand action can pair the eight electrons in four of the five \(d\) orbitals.
The resulting arrangement is:
\[
(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\ )
\]
The vacant inner \(d\) orbital combines with:
\[
1ns+2np
\]
The hybridisation is:
\[
\mathrm{dsp^2}
\]
Four equivalent hybrid orbitals are produced.
Since every \(d\) electron is paired, the standard complex is diamagnetic.
304. A four-coordinate \(d^8\) complex is found to be diamagnetic and to have all four ligands in one plane. The most suitable valence bond description is:
ⓐ. \(\mathrm{sp^3}\) hybridisation with two unpaired electrons
ⓑ. \(\mathrm{sp^3d^2}\) hybridisation with octahedral geometry
ⓒ. \(\mathrm{d^2sp^3}\) hybridisation with six ligand positions
ⓓ. \(\mathrm{dsp^2}\) hybridisation with square-planar geometry
Correct Answer: \(\mathrm{dsp^2}\) hybridisation with square-planar geometry
Explanation: Four-coordinate geometry may be tetrahedral or square planar, so coordination number alone is insufficient. The statement that all ligands lie in one plane identifies square-planar geometry. Diamagnetism shows that the \(d^8\) electrons are paired in the standard valence bond treatment. One vacant inner \(d\), one \(s\), and two \(p\) orbitals then form four \(\mathrm{dsp^2}\) hybrid orbitals. The magnetic and geometrical evidence therefore support the same assignment.
305. Study the proposed properties.
| Row | Feature | Square-planar \(\mathrm{dsp^2}\) complex |
|---|
| P | Number of hybrid orbitals | \(4\) |
| Q | Orbital set used | One \((n-1)d\), one \(ns\), and two \(np\) |
| R | Typical standard \(d^8\) magnetic behaviour | Diamagnetic |
| S | Ligand arrangement | Toward the vertices of a tetrahedron |
The inconsistent row is:
ⓐ. P
ⓑ. Q
ⓒ. S
ⓓ. R
Correct Answer: S
Explanation: Four orbitals participate in \(\mathrm{dsp^2}\) hybridisation. They consist of one inner \(d\), one \(s\), and two \(p\) orbitals. Their directions lie in one plane and point toward the corners of a square. Standard strongly paired \(d^8\) examples contain no unpaired electrons. Row S incorrectly assigns a tetrahedral orientation to the square-planar hybrid set.
306. Assertion: A standard square-planar \(d^8\) complex is often diamagnetic.
Reason: Electron pairing can place eight \(d\) electrons in four orbitals and leave one inner \(d\) orbital vacant for \(\mathrm{dsp^2}\) hybridisation.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: A \(d^8\) ion contains eight electrons in five \(d\) orbitals. Strong ligand influence can pair these electrons in four orbitals. The fifth inner \(d\) orbital then becomes available for \(\mathrm{dsp^2}\) hybridisation. All eight electrons are paired in this arrangement. The absence of unpaired electrons accounts for the diamagnetism stated in the Assertion.
307. The orbitals mixed during formation of four tetrahedrally directed hybrid orbitals are:
ⓐ. one \(s\), two \(p\), and one \(d\) orbital
ⓑ. two \(s\) and two \(p\) orbitals
ⓒ. one \(d\), one \(s\), and two \(p\) orbitals
ⓓ. one \(s\) and three \(p\) orbitals
Correct Answer: one \(s\) and three \(p\) orbitals
Explanation: The notation \(\mathrm{sp^3}\) represents one \(s\) orbital and three \(p\) orbitals. Their total number is four. Hybridisation produces four equivalent orbitals with tetrahedral orientation. No inner or outer \(d\) orbital is required. Each ligand donor atom occupies one of these four directions.
308. Tetrahedral complexes commonly retain a high-spin arrangement because their valence bond formation:
ⓐ. always requires complete pairing in the inner \(d\) orbitals
ⓑ. requires two vacant inner \(d\) orbitals
ⓒ. \(\mathrm{sp^3}\) needs no vacant inner \(d\) orbital
ⓓ. converts all metal electrons into ligand lone pairs
Correct Answer: \(\mathrm{sp^3}\) needs no vacant inner \(d\) orbital
Explanation: The tetrahedral hybrid set is formed only from one \(s\) and three \(p\) orbitals. Inner \(d\) orbitals are not needed for the bonding hybridisation. The metal therefore need not pair its \(d\) electrons to create a \(d\)-orbital vacancy. In many suitable cases, the electrons remain as unpaired as possible. This makes tetrahedral complexes commonly paramagnetic and high spin.
309. The ideal bond angle between two adjacent ligand directions in a regular tetrahedral complex is approximately:
ⓐ. \(90.0^\circ\)
ⓑ. \(109.5^\circ\)
ⓒ. \(120.0^\circ\)
ⓓ. \(180.0^\circ\)
Correct Answer: \(109.5^\circ\)
Explanation: Four equivalent \(\mathrm{sp^3}\) orbitals repel one another and adopt the most widely separated arrangement. Their directions point toward the vertices of a regular tetrahedron. The angle between any two ligand directions is approximately \(109.5^\circ\). Square-planar adjacent positions instead give \(90^\circ\). The tetrahedral arrangement has no ordinary pair of opposite ligand positions.
310. Consider the following statements about tetrahedral complexes in valence bond theory.
Statement I: Their hybridisation is \(\mathrm{sp^3}\).
Statement II: Four equivalent hybrid orbitals are formed.
Statement III: Inner-\(d\)-orbital pairing is compulsory before bonding.
Statement IV: Halide ligands commonly occur in tetrahedral examples.
The valid statements are:
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I, II, III and IV
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: Tetrahedral bonding is described by four \(\mathrm{sp^3}\) hybrid orbitals. These orbitals are equivalent and directed toward the vertices of a tetrahedron. Because no inner \(d\) orbital participates, compulsory \(d\)-electron pairing is unnecessary. Weak-field halide ligands commonly support tetrahedral high-spin examples. Statement III therefore gives an incorrect bonding requirement.
311. Study the high-spin tetrahedral electron counts.
| Row | Configuration | Proposed number of unpaired electrons |
|---|
| P | \(d^4\) | \(4\) |
| Q | \(d^5\) | \(5\) |
| R | \(d^6\) | \(4\) |
| S | \(d^8\) | \(0\) |
The inconsistent row is:
ⓐ. S
ⓑ. P
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: High-spin \(d^4\) has four singly occupied orbitals. High-spin \(d^5\) has five, while high-spin \(d^6\) has four after one electron pair forms. For \(d^8\), three orbitals are doubly occupied and two remain singly occupied. A tetrahedral \(d^8\) complex therefore has two unpaired electrons rather than zero. Row S incorrectly assigns diamagnetism to the high-spin arrangement.
312. Assertion: A tetrahedral \(d^8\) complex can be paramagnetic even though a square-planar \(d^8\) complex may be diamagnetic.
Reason: The tetrahedral \(\mathrm{sp^3}\) description does not require the pairing needed to make an inner \(d\) orbital vacant for square-planar \(\mathrm{dsp^2}\) hybridisation.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: In the square-planar case, the \(d^8\) electrons may pair in four inner \(d\) orbitals. This leaves no unpaired electrons and creates one vacant \(d\) orbital for \(\mathrm{dsp^2}\) hybridisation. In a tetrahedral complex, \(\mathrm{sp^3}\) hybridisation uses no \(d\) orbital. The \(d^8\) electrons can retain the high-spin arrangement with two unpaired electrons. The Reason therefore explains the magnetic contrast described in the Assertion.
313. A four-coordinate complex is known to be tetrahedral and high spin. Which property is least consistent with this information?
ⓐ. \(\mathrm{sp^3}\) hybridisation
ⓑ. All electrons paired in a \(d^8\) arrangement
ⓒ. No compulsory pairing of inner \(d\) electrons
ⓓ. Four equivalent ligand directions
Correct Answer: All electrons paired in a \(d^8\) arrangement
Explanation: Tetrahedral complexes are described by \(\mathrm{sp^3}\) hybridisation. Four equivalent orbitals point toward tetrahedral ligand positions. Since inner \(d\) orbitals do not participate, pairing to create a vacant inner \(d\) orbital is unnecessary. A tetrahedral high-spin \(d^8\) arrangement normally contains two unpaired electrons. Complete pairing is instead associated with the standard square-planar \(d^8\) case.
314. A four-coordinate \(d^8\) complex is experimentally diamagnetic. Another complex of the same metal ion is paramagnetic with two unpaired electrons. The most consistent geometries of the two complexes are respectively:
ⓐ. square planar and tetrahedral
ⓑ. tetrahedral and square planar
ⓒ. octahedral and linear
ⓓ. tetrahedral and octahedral
Correct Answer: square planar and tetrahedral
Explanation: A diamagnetic \(d^8\) four-coordinate complex requires all eight \(d\) electrons to be paired in the standard valence bond treatment. This is achieved through \(\mathrm{dsp^2}\) hybridisation and square-planar geometry. A tetrahedral \(d^8\) complex uses \(\mathrm{sp^3}\) hybridisation without compulsory \(d\)-electron pairing. Its high-spin arrangement contains two unpaired electrons. The contrasting magnetic observations therefore distinguish square-planar and tetrahedral geometries.
315. A six-coordinate \(d^6\) metal ion forms a complex in which all six \(d\) electrons are paired in three inner \(d\) orbitals. The predicted hybridisation and magnetic behaviour are:
ⓐ. \(\mathrm{sp^3d^2}\) and paramagnetic
ⓑ. \(\mathrm{sp^3}\) and diamagnetic
ⓒ. \(\mathrm{dsp^2}\) and paramagnetic
ⓓ. \(\mathrm{d^2sp^3}\) and diamagnetic
Correct Answer: \(\mathrm{d^2sp^3}\) and diamagnetic
Explanation: The metal-ion configuration is:
\[
d^6
\]
Pairing places the six electrons in three inner \(d\) orbitals:
\[
(\uparrow\downarrow)(\uparrow\downarrow)(\uparrow\downarrow)(\ )(\ )
\]
Two inner \(d\) orbitals are therefore vacant.
These combine with one \(s\) and three \(p\) orbitals:
\[
2(n-1)d+1ns+3np
\]
The hybridisation is:
\[
\mathrm{d^2sp^3}
\]
Six hybrid orbitals give octahedral geometry.
The number of unpaired electrons is:
\[
n=0
\]
The complex is therefore diamagnetic.
316. A student states that every complex containing \(\mathrm{NH_3}\) must be an inner-orbital complex. The best correction is:
ⓐ. \(\mathrm{NH_3}\) is never a ligand
ⓑ. the outcome depends on the metal, oxidation state, d-electron count, coordination number, and geometry
ⓒ. all ammine complexes are tetrahedral
ⓓ. \(\mathrm{NH_3}\) always acts as a weak-field anionic ligand
Correct Answer: the outcome depends on the metal, oxidation state, d-electron count, coordination number, and geometry
Explanation: Ammonia is a neutral ligand that donates through nitrogen. Its effect cannot be applied as an absolute rule to every metal ion. The oxidation state and \(d^n\) configuration influence whether pairing is required or energetically favourable. Geometry and coordination number must also be considered. Some cobalt(III) ammine complexes are paired inner-orbital complexes, but that does not justify a universal statement about all ammine compounds.
317. A weak-field ligand often leads to an octahedral outer-orbital complex because:
ⓐ. weak-field ligands always reduce the coordination number to \(4\)
ⓑ. two vacant \((n-1)d\) orbitals are automatically present for every metal ion
ⓒ. weak-field ligands force all inner \(d\) electrons to pair
ⓓ. without two vacant inner \(d\) orbitals, outer \(nd\) orbitals are used
Correct Answer: without two vacant inner \(d\) orbitals, outer \(nd\) orbitals are used
Explanation: An inner-orbital octahedral complex requires two vacant \((n-1)d\) orbitals. Weak-field ligands commonly do not produce the pairing needed to create those vacancies. The metal can then use two outer \(nd\) orbitals together with one \(s\) and three \(p\) orbitals. Valence bond theory describes this arrangement as \(\mathrm{sp^3d^2}\) hybridisation. Retention of more unpaired electrons is therefore linked with the outer-orbital description.
318. The usual valence bond reasoning for a strong-field octahedral complex follows the sequence:
ⓐ. Strong field \(\rightarrow\) pairing \(\rightarrow\) vacant inner \(d\) orbitals \(\rightarrow\) \(\mathrm{d^2sp^3}\)
ⓑ. Strong ligand influence \(\rightarrow\) loss of all \(d\) electrons \(\rightarrow\) \(\mathrm{sp^3}\) hybridisation
ⓒ. Strong ligand influence \(\rightarrow\) maximum unpaired electrons \(\rightarrow\) \(\mathrm{sp^3d^2}\) hybridisation
ⓓ. Strong ligand influence \(\rightarrow\) coordination number \(2\) \(\rightarrow\) linear geometry
Correct Answer: Strong field \(\rightarrow\) pairing \(\rightarrow\) vacant inner \(d\) orbitals \(\rightarrow\) \(\mathrm{d^2sp^3}\)
Explanation: A strong ligand field may cause electrons to pair in fewer inner \(d\) orbitals. Pairing can release two \((n-1)d\) orbitals for bonding. These orbitals mix with one \(s\) and three \(p\) orbitals. The resulting \(\mathrm{d^2sp^3}\) set contains six octahedrally directed hybrid orbitals. This sequence connects ligand influence, electron arrangement, hybridisation, and geometry.
319. Study the ligand-effect descriptions.
| Row | Condition | Likely consequence in a suitable octahedral complex |
|---|
| P | Strong-field ligand | Pairing may increase |
| Q | Weak-field ligand | More unpaired electrons may remain |
| R | Two vacant inner \(d\) orbitals available | \(\mathrm{d^2sp^3}\) may form |
| S | Inner \(d\) electrons remain maximally unpaired | Diamagnetism is guaranteed |
The inconsistent row is:
ⓐ. P
ⓑ. Q
ⓒ. S
ⓓ. R
Correct Answer: S
Explanation: Strong-field ligands may promote pairing, while weak-field ligands often preserve more unpaired electrons. Availability of two inner \(d\) orbitals supports \(\mathrm{d^2sp^3}\) hybridisation. Retaining the maximum number of unpaired electrons produces paramagnetism rather than guaranteed diamagnetism. The exact unpaired-electron count depends on the \(d^n\) configuration. Row S therefore reverses the magnetic consequence.
320. In \(\mathrm{[Co(CN)_6]^{3-}}\), the oxidation state and \(d\)-electron configuration of cobalt are respectively:
ⓐ. \(+2\) and \(d^7\)
ⓑ. \(+3\) and \(d^6\)
ⓒ. \(+3\) and \(d^4\)
ⓓ. \(0\) and \(d^9\)
Correct Answer: \(+3\) and \(d^6\)
Explanation: Let the oxidation state of cobalt be:
\[
x
\]
Each cyanido ligand has charge:
\[
-1
\]
The charge-balance equation is:
\[
x+6(-1)=-3
\]
Simplifying:
\[
x-6=-3
\]
Hence:
\[
x=+3
\]
Neutral cobalt is:
\[
\mathrm{[Ar]\,3d^7\,4s^2}
\]
Therefore, \(\mathrm{Co^{3+}}\) is:
\[
\mathrm{3d^6}
\]
The complex contains a cobalt(III) \(d^6\) centre.
