Question:
Which of the following does not represent property stated against it?
Which of the following does not represent property stated against it?
Updated On: Nov 14, 2025
- $CO ^{+2}< Fe ^{+2}< Mn ^{+2}$ - lonic size
- $Ti < V < Mn $ - Number of oxidation states
- $Cr ^{+2}< Mn ^{+2}< Fe ^{+2}$ - Paramagnetic behaviour
- none of the above
Show Solution
Verified By Collegedunia
The Correct Option is D
Approach Solution - 1
To address the question of which statement does not represent the property stated against it, we need to analyze each provided option in terms of the concepts involved:
- Ionic Size: The given order is \(CO^{+2} < Fe^{+2} < Mn^{+2}\). Ionic size generally decreases with an increase in nuclear charge, i.e., with a decrease in the number of remaining electrons for similar ions (those having similar electric charges).
Reasoning: All given ions have a similar charge, \(+2\). As we move from Mn to Co across the period in the periodic table, the nuclear charge increases while the electron number remains relatively constant, pulling electrons closer and decreasing ionic size.
Hence, this is correct. - Number of Oxidation States: The order given is \(Ti < V < Mn\).
Explanation: Transition elements are known for having multiple oxidation states. Titanium (Ti) generally has +2, +3, +4; Vanadium (V) has +2, +3, +4, +5; and Manganese (Mn) has several from +2 to +7.
Hence, this order accurately depicts the number of oxidation states. - Paramagnetic Behaviour: The provided order is \(Cr^{+2} < Mn^{+2} < Fe^{+2}\).
Explanation: Paramagnetism arises from unpaired electrons. Upon measuring the number of unpaired electrons: Cr has four, Mn has five, and Fe has six unpaired d-electrons, adhering to Hund's rule where more unpaired electrons tend to enhance paramagnetic behavior.
Thus, this order on paramagnetic behavior is maintained as given.
Based on the considerations described, all given options accurately depict the properties stated. Therefore, the correct response is none of the above.
Was this answer helpful?
0
0
Show Solution
Verified By Collegedunia
Approach Solution -2
$\underset{24}{ Cr }[ Ar ] 3 d ^{5} 4 s ^{1} \,\,\, Cr ^{2+} 3 d ^{4} 4 s ^{0}=4$ unpaired electrons
$Mn [ Ar ] 3 d ^{5} 4 s ^{2} Mn ^{2+} 3 d ^{5} 4 s ^{0}=5$ unpairedelectrons
$Fe [ Ar ] 3 d ^{6} 4 s ^{2} \,\,\,Fe ^{2+} 3 d ^{6} 4 s ^{0}=4$ unpaired electrons
Since both $Cr ^{2+}$ and $Fe ^{2+}$ contain $4$ unpaired electrons, the correct option is $C$.
Option D is also correct because density order given is not correct.
$Mn [ Ar ] 3 d ^{5} 4 s ^{2} Mn ^{2+} 3 d ^{5} 4 s ^{0}=5$ unpairedelectrons
$Fe [ Ar ] 3 d ^{6} 4 s ^{2} \,\,\,Fe ^{2+} 3 d ^{6} 4 s ^{0}=4$ unpaired electrons
Since both $Cr ^{2+}$ and $Fe ^{2+}$ contain $4$ unpaired electrons, the correct option is $C$.
Option D is also correct because density order given is not correct.
Was this answer helpful?
0
0
Top KCET Chemistry Questions
- Which of the following is not true for oxidation?
- When $50\, cm ^{3}$ of $0.2\, N$ $H_{2} SO _{4}$ is mixed with $50 cm ^{3}$ of $1 N\, KOH$, the heat liberated is :
- Which of the following is NOT a pair of functional isomers?
- The vapour pressure of due to
- A nitrogen containing organic compound gave an oily liquid on heating with bromine and potassium hydroxide solution. On shaking the product with acetic anhydride, an antipyretic drug was obtained. The reactions indicate that the starting compound is :
View More Questions
Top KCET properties of d block elements Questions
Top KCET Questions
- Which of the following is not true for oxidation?
- If $a, b$ and $c \in N$ which one of the following is not true ?
- When $50\, cm ^{3}$ of $0.2\, N$ $H_{2} SO _{4}$ is mixed with $50 cm ^{3}$ of $1 N\, KOH$, the heat liberated is :
- If $a | (b + c)$ and $a| (b-c)$ where $a,b,c \in \, N $ then
- Which of the following is NOT a pair of functional isomers?
View More Questions
Concepts Used:
Properties of D Block Elements
- Multiple oxidation states- The oxidation states of d block elements show very few energy gaps; therefore, they exhibit many oxidation states. Also, the energy difference between s and d orbital is very less. Therefore both the electrons are involved in ionic and covalent bond formation, which ultimately leads to multiple oxidation states.
- Formation of complex compounds- Ligands show a binding behavior and can form so many stable complexes with the help of transition metals. This property is mainly due to:
- Availability of vacant d orbitals.
- Comparatively small sizes of metals.
- Hardness- Transition elements are tough and have high densities because of the presence of unpaired electrons.
- Melting and boiling points- Melting and boiling points of transition are very high because of the presence of unpaired electrons and partially filled d orbitals. They form strong bonds and have high melting and boiling points.
- Atomic radii- The atomic and ionic radius of the transition elements decreases as we move from Group 3 to group 6. However, it remains the same between group 7 and group 10, and from group 11 to group 12 increases.
- Ionization enthalpy- The ionization enthalpies of the transition elements are generally on the greater side as compared to the S block elements