MCQ
Which have melting point below $500\,^{\circ} C$.
- ✓$Zn ,\, Cd$
- B$Ag , \,Cu$
- C$Cd , \,Cu$
- D$Ag , \,Zn$
✓
Answer
Correct option: A.
$Zn ,\, Cd$
a
Zinc and cadmium metal are soft metals. They form weak metallic bond. The melting point of zinc is $146\,^{\circ} C$ and melting point of $Cd$ is $321\,^{\circ} C$.
Zinc and cadmium metal are soft metals. They form weak metallic bond. The melting point of zinc is $146\,^{\circ} C$ and melting point of $Cd$ is $321\,^{\circ} C$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Consider the complex $[Co(NH_3)_4 CO_3]ClO_4$ , in which coordination number, oxidation number and number of $d$-electrons on the metal are respectively.
→Bond length of ethane $(I),$ ethene $(II),$ acetylene $(III)$ and benzene $(IV)$ follows the order
→$C{H_3}MgI$ will give methane with
→Identify the no. of $\alpha -\,$ Hydrogen in the given compound?
→Ammonia can be dried by
Assertion : Zinc can be used while copper cannot be used in the recovery of $Ag$ from the complex $[Ag(CN)_2]^-$.
→Reason : Zinc is a powerful reducing agent than copper.
For the first order decompsition reaction of $N_2O_5$, it is found that -
→$(a)$ $2N_2O_5\rightarrow\,\,4NO_2(g)+O_2(g)-\frac{d[N_2O_5]}{dt}=k[N_2O_5]$
$(a)$ $N_2O_5\rightarrow\,\,2NO_2(g)+1/2\,\,O_2(g)-\frac{d[N_2O_5]}{dt}=k'[N_2O_5]$
which of the following is true ?
Bond angle between two hybrid orbitals is ${105^o}.\% $ $s-$orbital character of hybrid orbital is
→Which of the following compound gives the smell of mustard oil:
$1 \times 10^{-5}\,M \textrm {AgNO } _ { 3 }$ is added to $1\,L$ of saturated solution of $AgBr$. The conductivity of this solution at $298\,K$ is $.........\times 10^{-8}\,S\,m ^{-1}$
→$\left[\right.$ Given : $K _{ sp }( AgBr )=4.9 \times 10^{-13}$ at $298 K$
$\lambda_{ Ag ^{+}}^0=6 \times 10^{-3} Sm ^2\,mol ^{-1}$
$\lambda_{ Br ^{-}}^0=8 \times 10^{-3} Sm ^2\,mol ^{-1}$
$\left.\lambda_{ NO _3^{-}}^0=7 \times 10^{-3} Sm ^2\,mol ^{-1}\right]$