Advertisements
Advertisements
प्रश्न
Assertion (A): Transition metals have high enthalpy of atomisation.
Reason (R): Greater number of unpaired electrons in transition metals results in weak metallic bonding.
पर्याय
Both (A) and (R) are true and (R) is the correct explanation of (A).
Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(A) is true, but (R) is false.
(A) is false, but (R) is true.
उत्तर
(A) is true, but (R) is false.
Explanation:
The assertion (A) is true: Transition metals exhibit a high enthalpy of atomization. This is owing to the strong metallic bonds found in these metals.
The reasoning (R) is false: The existence of unpaired electrons in transition metals causes stronger, rather than weaker, metallic bonding. The delocalization of d and s electrons across the metal lattice helps to form strong metallic connections.
APPEARS IN
संबंधित प्रश्न
Why +2 oxidation state of manganese is more stable?
Calculate the ‘spin only’ magnetic moment of \[\ce{M^{2+}_{ (aq)}}\] ion (Z = 27).
Calculate the number of unpaired electrons in the following gaseous ions:
Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in an aqueous solution?
EΘ of Cu is + 0.34V while that of Zn is – 0.76V. Explain.
Answer the following question:
Which element of the first transition series has highest second ionisation enthalpy?
A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. \[\ce{H2SO4}\] and \[\ce{NaCl}\], chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.
How is the variability in oxidation states of transition metals different from that of p-block elements?
The given graph shows the trends in melting points of transition metals:
Explain the reason why Cr has the highest melting point and manganese (Mn) has a lower melting point.
Consider the following standard electrode potential values:
\[\ce{Sn^{2+}_{ (aq)} + 2e^- -> Sn_{(s)}}\]; E0 = −0.14 V
\[\ce{Fe^{3+}_{ (aq)} + e^- -> Fe^{2+}_{ (aq)}}\]; E0 = +0.77 V
What is the cell reaction and potential for the spontaneous reaction that occurs?
In order to protect iron from corrosion, which one will you prefer as a sacrificial electrode, Ni or Zn? Why? (Given standard electrode potentials of Ni, Fe and Zn are -0.25 V, -0.44 V and -0.76 V respectively.)
