Question

Which of the following does not represent property stated against it?

• A. $CO ^{+2}< Fe ^{+2}< Mn ^{+2}$ - lonic size
• B. $Ti < V < Mn $ - Number of oxidation states
• C. $Cr ^{+2}< Mn ^{+2}< Fe ^{+2}$ - Paramagnetic behaviour
• D. none of the above

Answer 1

The Correct Option is D

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: 

1. 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.
2. 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.
3. 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.

Answer 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.