
Which one of the following does not correctly represent the correct order of the property indicated against it?
A. $M{{n}^{3+}}(22)=[Ar]3{{d}^{4}}$; increasing magnetic moment
B. $Ti < V < Cr < Mn$; increasing melting point
C. $Ti < V < Mn < Cr$; increasing 2nd ionization enthalpy
D. $Ti < V < Cr < Mn$; increasing number of oxidation states
Answer
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Hint: IUPAC defines transition elements as an element having a d subshell that is partially filled with electrons, or an element that has the ability to form stable cations with an incompletely filled d orbital. All the elements shown in the question are in the category of transition elements.
Complete answer: Any element which corresponds to the d-block of the modern periodic table is considered to be a transition element. General electronic configurations corresponding to transition metals are $(n-1){{d}^{0-10}}n{{s}^{1-2}}$. Now we will discuss all the options
A. $T{{i}^{3+}}<{{V}^{3+}}Magnetic moment is generally dependent on the unpaired electrons, higher the number of unpaired electron higher is the magnetic moment so the unpaired electron in
$T{{i}^{3+}}(19)=[Ar]3{{d}^{1}}$; unpaired electron = 1
${{V}^{3+}}(20)=[Ar]3{{d}^{2}}$; unpaired electron = 2
$C{{r}^{3+}}(21)=[Ar]3{{d}^{3}}$; unpaired electron = 3
$M{{n}^{3+}}(22)=[Ar]3{{d}^{4}}$; unpaired electron = 4
Hence the option A is correct.
B. $Ti < V < Cr < Mn$; increasing melting point
Melting point decreases as we move across the period i.e. from left to right according to this concept melting point of Mn should be less so option B is not correct.
C. $Ti < V < Mn < Cr$; increasing 2nd ionization enthalpy
Ionization enthalpy increases as we move from left to right across the period, So option C is also correct.
D. $Ti < V < Cr < Mn $; increasing number of oxidation states
Talking about oxidation states, titanium has 3, vanadium has 4, chromium has 5 and manganese has 6 oxidation states. So we can say that option D is also correct.
Hence according to question, B does not correctly represent the correct order of the property indicated against it.
Note:
Ionization enthalpy refers to the amount of energy that must be supplied to an element for the removal of a valence electron. The greater the effective nuclear charge acting on the electrons, the greater the ionization potential of the element and effective nuclear charge increases as we move from left to right.
Complete answer: Any element which corresponds to the d-block of the modern periodic table is considered to be a transition element. General electronic configurations corresponding to transition metals are $(n-1){{d}^{0-10}}n{{s}^{1-2}}$. Now we will discuss all the options
A. $T{{i}^{3+}}<{{V}^{3+}}
$T{{i}^{3+}}(19)=[Ar]3{{d}^{1}}$; unpaired electron = 1
${{V}^{3+}}(20)=[Ar]3{{d}^{2}}$; unpaired electron = 2
$C{{r}^{3+}}(21)=[Ar]3{{d}^{3}}$; unpaired electron = 3
$M{{n}^{3+}}(22)=[Ar]3{{d}^{4}}$; unpaired electron = 4
Hence the option A is correct.
B. $Ti < V < Cr < Mn$; increasing melting point
Melting point decreases as we move across the period i.e. from left to right according to this concept melting point of Mn should be less so option B is not correct.
C. $Ti < V < Mn < Cr$; increasing 2nd ionization enthalpy
Ionization enthalpy increases as we move from left to right across the period, So option C is also correct.
D. $Ti < V < Cr < Mn $; increasing number of oxidation states
Talking about oxidation states, titanium has 3, vanadium has 4, chromium has 5 and manganese has 6 oxidation states. So we can say that option D is also correct.
Hence according to question, B does not correctly represent the correct order of the property indicated against it.
Note:
Ionization enthalpy refers to the amount of energy that must be supplied to an element for the removal of a valence electron. The greater the effective nuclear charge acting on the electrons, the greater the ionization potential of the element and effective nuclear charge increases as we move from left to right.
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