201. Why was a bonding theory needed for coordination compounds beyond writing only their formulas?
ⓐ. Because formulas can show colour directly but not composition
ⓑ. Formulas alone do not explain geometry, magnetism, and bonding
ⓒ. Because formulas do not show the oxidation state of the metal at all
ⓓ. Because formulas cannot distinguish ligands from counter ions
Correct Answer: Formulas alone do not explain geometry, magnetism, and bonding
Explanation: A formula can tell which metal and ligands are present, but it does not by itself explain how those ligands are arranged in space. It also does not fully account for why some complexes are paramagnetic while others are diamagnetic. To understand shape and magnetic behaviour, a bonding model is needed. This is why theories such as valence bond theory were introduced for coordination compounds.
202. In valence bond theory, the bond between a ligand and the metal is formed by:
ⓐ. donation of a lone pair from the ligand into an empty orbital of the metal
ⓑ. transfer of an electron from the metal to the ligand without any sharing
ⓒ. overlap of two half-filled ligand orbitals with no metal participation
ⓓ. attraction between only outer-sphere ions and solvent molecules
Correct Answer: donation of a lone pair from the ligand into an empty orbital of the metal
Explanation: Valence bond theory treats ligands as electron-pair donors and the metal as an electron-pair acceptor. The ligand supplies a lone pair into a vacant orbital on the metal ion or atom. This produces a coordinate bond. The idea is consistent with the Lewis base role of the ligand and the Lewis acid role of the metal centre.
203. Which statement correctly describes the role of hybridisation in valence bond theory for coordination compounds?
ⓐ. Hybridisation is used only to calculate oxidation state
ⓑ. Hybridisation is used only for naming the ligands
ⓒ. Hybridisation is used only to count outer-sphere ions
ⓓ. It explains orbital arrangement and geometry
Correct Answer: It explains orbital arrangement and geometry
Explanation: In valence bond theory, suitable atomic orbitals of the metal mix to form hybrid orbitals. These hybrid orbitals are then directed in specific spatial orientations, which helps explain the geometry of the complex. For example, different hybridisations are associated with octahedral, tetrahedral, or square planar structures. So hybridisation connects bonding with geometry.
204. Which of the following pairs is correctly matched in valence bond terms?
ⓐ. octahedral complex — \(sp^3\)
ⓑ. tetrahedral complex — \(sp^3\)
ⓒ. square planar complex — \(sp^3d^2\)
ⓓ. octahedral complex — \(dsp^2\)
Correct Answer: tetrahedral complex — \(sp^3\)
Explanation: In valence bond treatment, tetrahedral complexes are described using \(sp^3\) hybridisation. Octahedral complexes are associated with six hybrid orbitals, usually described as either \(d^2sp^3\) or \(sp^3d^2\). Square planar complexes are commonly described by \(dsp^2\) hybridisation. So only the tetrahedral–\(sp^3\) match is correct here.
205. The term inner orbital complex in valence bond theory refers to an octahedral complex that uses:
ⓐ. \((n-1)d\) orbitals in hybridisation
ⓑ. only \(ns\) orbitals in bonding
ⓒ. only outer \(nd\) orbitals in hybridisation
ⓓ. no \(d\)-orbital participation at all
Correct Answer: \((n-1)d\) orbitals in hybridisation
Explanation: An inner orbital complex uses the inner \((n-1)d\) orbitals of the metal during hybridisation. In octahedral complexes, this leads to \(d^2sp^3\) hybridisation in the valence bond picture. Such complexes often arise when electron pairing occurs in the \((n-1)d\) set. The term inner orbital therefore refers to use of the inner \(d\)-orbitals rather than the outer \(nd\) orbitals.
206. The term outer orbital complex in valence bond theory refers to an octahedral complex that uses:
ⓐ. only \(4s\) and \(4p\) orbitals
ⓑ. only \((n-1)d\) orbitals
ⓒ. outer \(nd\) orbitals along with \(ns\) and \(np\) orbitals
ⓓ. ligand orbitals only, with no metal orbital mixing
Correct Answer: outer \(nd\) orbitals along with \(ns\) and \(np\) orbitals
Explanation: In an outer orbital octahedral complex, the metal uses the outer \(nd\) orbitals together with \(ns\) and \(np\) orbitals. In valence bond language, this corresponds to \(sp^3d^2\) hybridisation. Such a description is used when pairing in the inner \((n-1)d\) orbitals does not take place. So the distinction between inner and outer orbital complexes depends on which set of \(d\)-orbitals participates.
207. Which hybridisation is associated with an inner orbital octahedral complex in valence bond theory?
ⓐ. \(sp^3\)
ⓑ. \(dsp^2\)
ⓒ. \(sp^3d^2\)
ⓓ. \(d^2sp^3\)
Correct Answer: \(d^2sp^3\)
Explanation: An inner orbital octahedral complex uses two inner \((n-1)d\) orbitals, one \(s\) orbital, and three \(p\) orbitals. These combine to give \(d^2sp^3\) hybridisation. The resulting six hybrid orbitals point toward the corners of an octahedron. This is the usual valence bond description for inner orbital octahedral complexes.
208. Which hybridisation is associated with an outer orbital octahedral complex in valence bond theory?
ⓐ. \(dsp^2\)
ⓑ. \(sp^3d^2\)
ⓒ. \(sp^3\)
ⓓ. \(d^2sp^3\)
Correct Answer: \(sp^3d^2\)
Explanation: In an outer orbital octahedral complex, the metal uses one \(s\), three \(p\), and two outer \(d\) orbitals. These mix to form six equivalent \(sp^3d^2\) hybrid orbitals. Those orbitals are directed octahedrally around the metal. This is why \(sp^3d^2\) is the valence bond description for outer orbital octahedral complexes.
209. In an inner orbital octahedral complex of a first-row transition metal, which set of metal orbitals is used to form the six hybrid orbitals?
ⓐ. one \(ns\), three \(np\), and two outer \(nd\) orbitals
ⓑ. two inner \((n-1)d\), one \(ns\), and three \(np\) orbitals
ⓒ. four \(ns\) orbitals and two \(np\) orbitals
ⓓ. six ligand orbitals only, with no metal orbital participation
Correct Answer: two inner \((n-1)d\), one \(ns\), and three \(np\) orbitals
Explanation: An inner orbital octahedral complex is described by \(d^2sp^3\) hybridisation. The two \(d\)-orbitals used in this notation come from the inner \((n-1)d\) level of the metal ion. They combine with one \(ns\) orbital and three \(np\) orbitals to give six hybrid orbitals directed octahedrally. This description is different from an outer orbital complex, where the outer \(nd\) orbitals are used instead. Therefore the correct orbital set is two inner \((n-1)d\), one \(ns\), and three \(np\) orbitals.
210. In valence bond theory, why is an outer orbital octahedral complex represented by \(sp^3d^2\) rather than \(d^2sp^3\)?
ⓐ. It uses outer \(nd\), not inner \((n-1)d\), orbitals
ⓑ. It has only four coordinate bonds around the metal
ⓒ. It contains no \(d\)-orbital contribution in bonding
ⓓ. It must always be square planar and diamagnetic
Correct Answer: It uses outer \(nd\), not inner \((n-1)d\), orbitals
Explanation: The order in the hybridisation notation indicates which set of orbitals is being used in the standard valence bond description. In an outer orbital octahedral complex, the metal uses one \(s\), three \(p\), and two outer \(d\)-orbitals of the same principal shell. This gives the notation \(sp^3d^2\). The notation \(d^2sp^3\) is reserved for inner orbital octahedral complexes, where two inner \((n-1)d\) orbitals participate. Thus the distinguishing point is the use of outer \(nd\) orbitals.
211. According to valence bond theory, what is the most appropriate description of \([Fe(CN)_6]^{4-}\)?
ⓐ. outer orbital, \(sp^3d^2\), paramagnetic
ⓑ. inner orbital, \(d^2sp^3\), diamagnetic
ⓒ. tetrahedral, \(sp^3\), paramagnetic
ⓓ. square planar, \(dsp^2\), diamagnetic
Correct Answer: inner orbital, \(d^2sp^3\), diamagnetic
Explanation: \(\textbf{Given:}\)
Complex \([Fe(CN)_6]^{4-}\)
Ligand \(CN^-\) is a strong-field ligand at this level
\(\textbf{Required:}\)
VBT description of the complex
\(\textbf{Relevant Principle:}\)
Strong-field ligands can cause pairing of electrons in the inner \((n-1)d\) orbitals, allowing formation of an inner orbital octahedral complex.
\(\textbf{Find the oxidation state of iron:}\)
Let oxidation state of iron be \(x\).
\[x + 6(-1) = -4\]
\[x = +2\]
\(\textbf{Identify the }d\textbf{ count:}\)
\(Fe^{2+}\) is \(3d^6\).
\(\textbf{Apply ligand effect:}\)
Because \(CN^-\) is strong field, pairing occurs in the \(3d\) set.
Two inner \(3d\) orbitals become available for hybridisation.
\(\textbf{Hybridisation obtained:}\)
\[d^2sp^3\]
\(\textbf{Magnetic consequence:}\)
All electrons become paired in this description, so the complex is diamagnetic.
\(\textbf{Final Answer:}\)
\([Fe(CN)_6]^{4-}\) is an inner orbital, \(d^2sp^3\), diamagnetic complex.
212. According to valence bond theory, what is the most appropriate description of \([FeF_6]^{4-}\)?
ⓐ. inner orbital, \(d^2sp^3\), diamagnetic
ⓑ. square planar, \(dsp^2\), diamagnetic
ⓒ. tetrahedral, \(sp^3\), paramagnetic
ⓓ. outer orbital, \(sp^3d^2\), paramagnetic
Correct Answer: outer orbital, \(sp^3d^2\), paramagnetic
Explanation: \(\textbf{Given:}\)
Complex \([FeF_6]^{4-}\)
Ligand \(F^-\) is a weak-field ligand at this level
\(\textbf{Required:}\)
VBT description of the complex
\(\textbf{Relevant Principle:}\)
Weak-field ligands generally do not cause pairing of electrons in the inner \((n-1)d\) orbitals, so octahedral bonding uses outer \(d\)-orbitals.
\(\textbf{Find the oxidation state of iron:}\)
Let oxidation state of iron be \(x\).
\[x + 6(-1) = -4\]
\[x = +2\]
\(\textbf{Identify the }d\textbf{ count:}\)
\(Fe^{2+}\) is \(3d^6\).
\(\textbf{Apply ligand effect:}\)
Since \(F^-\) is weak field, pairing is not favoured in the inner \(3d\) orbitals.
So the complex uses outer orbitals for octahedral hybridisation.
\(\textbf{Hybridisation obtained:}\)
\[sp^3d^2\]
\(\textbf{Magnetic consequence:}\)
Unpaired electrons remain, so the complex is paramagnetic.
\(\textbf{Final Answer:}\)
\([FeF_6]^{4-}\) is an outer orbital, \(sp^3d^2\), paramagnetic complex.
213. According to valence bond theory, tetrahedral complexes are generally:
ⓐ. \(dsp^2\) and usually diamagnetic
ⓑ. \(sp^3\) and usually paramagnetic
ⓒ. \(d^2sp^3\) and always diamagnetic
ⓓ. \(sp^3d^2\) and always diamagnetic
Correct Answer: \(sp^3\) and usually paramagnetic
Explanation: Tetrahedral complexes are described by \(sp^3\) hybridisation in valence bond theory. In the common valence-bond treatment, they are usually considered high-spin and therefore often paramagnetic. This is because electron pairing is generally not favoured in the situations that commonly give tetrahedral complexes. So \(sp^3\) geometry and paramagnetic behaviour are the usual association.
214. Which statement best describes \([Ni(CN)_4]^{2-}\) in valence bond theory?
ⓐ. It is a square planar \(dsp^2\) complex and is diamagnetic
ⓑ. It is a tetrahedral \(sp^3\) complex and is paramagnetic
ⓒ. It is an octahedral \(sp^3d^2\) complex and is paramagnetic
ⓓ. It is an octahedral \(d^2sp^3\) complex and is diamagnetic
Correct Answer: It is a square planar \(dsp^2\) complex and is diamagnetic
Explanation: For \(Ni^{2+}\), the electronic situation is \(3d^8\). The ligand \(CN^-\) is strong field, so pairing is favoured in the \(3d\) orbitals. This allows one \(d\), one \(s\), and two \(p\) orbitals to hybridise as \(dsp^2\). The result is a square planar complex with all electrons paired, so it is diamagnetic.
215. Which comparison between \([NiCl_4]^{2-}\) and \([Ni(CN)_4]^{2-}\) is correct?
ⓐ. Both are square planar and diamagnetic
ⓑ. Both are tetrahedral and paramagnetic
ⓒ. \([NiCl_4]^{2-}\) is square planar and diamagnetic, while \([Ni(CN)_4]^{2-}\) is tetrahedral and paramagnetic
ⓓ. \([NiCl_4]^{2-}\) is tetrahedral and paramagnetic, while \([Ni(CN)_4]^{2-}\) is square planar and diamagnetic
Correct Answer: \([NiCl_4]^{2-}\) is tetrahedral and paramagnetic, while \([Ni(CN)_4]^{2-}\) is square planar and diamagnetic
Explanation: The ligand \(Cl^-\) is weak field, so \([NiCl_4]^{2-}\) is treated as a tetrahedral \(sp^3\) complex and remains paramagnetic. The ligand \(CN^-\) is strong field, so \([Ni(CN)_4]^{2-}\) is described as square planar with \(dsp^2\) hybridisation and paired electrons. The same metal ion can therefore show different geometry and magnetic behaviour depending on the ligand. This comparison is one of the standard illustrations of ligand effect in valence bond treatment.
216. Which statement about magnetic behaviour in valence bond theory is correct?
ⓐ. A complex is diamagnetic only when its metal has oxidation state zero
ⓑ. A complex is paramagnetic only when it contains counter ions outside the bracket
ⓒ. It depends on whether electrons remain unpaired
ⓓ. Magnetic behaviour depends only on coordination number and not on electron pairing
Correct Answer: It depends on whether electrons remain unpaired
Explanation: Valence bond theory relates magnetism to the presence or absence of unpaired electrons in the metal after bonding is considered. If unpaired electrons remain, the complex is paramagnetic. If all electrons become paired, the complex is diamagnetic. So the key issue is electron pairing, not just oxidation state, counter ions, or coordination number alone.
217. Which valence bond description is most appropriate for \([PtCl_4]^{2-}\)?
ⓐ. square planar, \(dsp^2\), diamagnetic
ⓑ. tetrahedral, \(sp^3\), paramagnetic
ⓒ. octahedral, \(sp^3d^2\), diamagnetic
ⓓ. octahedral, \(d^2sp^3\), paramagnetic
Correct Answer: square planar, \(dsp^2\), diamagnetic
Explanation: Platinum in \([PtCl_4]^{2-}\) is in the \(+2\) oxidation state and belongs to the \(d^8\) class. In valence bond treatment, such four-coordinate complexes are commonly described as square planar with \(dsp^2\) hybridisation. This arrangement is associated with electron pairing and diamagnetic behaviour. So the complex is treated as square planar, \(dsp^2\), and diamagnetic.
218. Which statement about \([NiCl_4]^{2-}\) is correct in valence bond theory?
ⓐ. It is square planar and diamagnetic
ⓑ. It is octahedral and diamagnetic
ⓒ. It is tetrahedral and paramagnetic
ⓓ. It is square planar and paramagnetic
Correct Answer: It is tetrahedral and paramagnetic
Explanation: The ligand \(Cl^-\) is weak field at this level, so it does not favour pairing strongly in \(Ni^{2+}\). As a result, \([NiCl_4]^{2-}\) is treated as a tetrahedral complex with \(sp^3\) hybridisation. Unpaired electrons remain in the metal ion, so the complex is paramagnetic. This makes it different from \([Ni(CN)_4]^{2-}\), which is square planar and diamagnetic.
219. Which comparison between \([Fe(CN)_6]^{4-}\) and \([FeF_6]^{4-}\) is correct in valence bond theory?
ⓐ. \([Fe(CN)_6]^{4-}\) and \([FeF_6]^{4-}\) are both inner orbital and diamagnetic
ⓑ. \([Fe(CN)_6]^{4-}\) and \([FeF_6]^{4-}\) are both outer orbital and paramagnetic
ⓒ. \([Fe(CN)_6]^{4-}\) is outer orbital, while \([FeF_6]^{4-}\) is inner orbital
ⓓ. \([Fe(CN)_6]^{4-}\) is inner orbital; \([FeF_6]^{4-}\) is outer orbital
Correct Answer: \([Fe(CN)_6]^{4-}\) is inner orbital; \([FeF_6]^{4-}\) is outer orbital
Explanation: Both complexes contain \(Fe^{2+}\), so the metal has the same \(d^6\) count in each case. The difference comes from ligand strength: \(CN^-\) is strong field and favours pairing, while \(F^-\) is weak field and does not. This makes \([Fe(CN)_6]^{4-}\) an inner orbital \(d^2sp^3\) diamagnetic complex, whereas \([FeF_6]^{4-}\) is an outer orbital \(sp^3d^2\) paramagnetic complex. The example is a standard comparison showing how ligands affect pairing and magnetism.
220. How many unpaired electrons are expected in \([FeF_6]^{4-}\) according to valence bond theory?
ⓐ. \(0\)
ⓑ. \(4\)
ⓒ. \(2\)
ⓓ. \(6\)
Correct Answer: \(4\)
Explanation: \(\textbf{Given:}\)
Complex \([FeF_6]^{4-}\)
Ligand \(F^-\) is weak field
\(\textbf{Required:}\)
Number of unpaired electrons in the complex
\(\textbf{Relevant Principle:}\)
Weak-field ligands usually do not cause pairing in the inner \((n-1)d\) orbitals, so a high-spin arrangement is retained.
\(\textbf{Find the oxidation state of iron:}\)
Let oxidation state of iron be \(x\).
\[x + 6(-1) = -4\]
\[x = +2\]
\(\textbf{Identify the }d\textbf{ count:}\)
\(Fe^{2+}\) is \(3d^6\).
\(\textbf{Apply weak-field behaviour:}\)
For high-spin \(d^6\), electrons remain as unpaired as possible before pairing.
This gives \(4\) unpaired electrons.
\(\textbf{Final Answer:}\)
Number of unpaired electrons \(= 4\)
