If in forsterite mineral- bivalent Mg2- ions are to be replaced by unipositive Na- ions- and if Na- ions are occupying half of the TVs in fcc lattice- the arrangement of rest of the constituents is kept same- then the formula of the new solid is-

The correct option is C Na_4SiO_4 Number of nbsp; Na^⊕ nbsp;ions in fcc = 1/2× TV=1/2× 8 = 4/unit cellSo, formula of solid nbsp; = n ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Oxidation States

If in forster...

Question

If in forsterite mineral, bivalent $M g^{2 +}$ ions are to be replaced by unipositive $N a^{\oplus}$ ions, and if $N a^{\oplus}$ ions are occupying half of the TVs in fcc lattice, the arrangement of rest of the constituents is kept same, then the formula of the new solid is:

N a_{2} S i O_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N a_{2} S i O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N a_{4} S i O_{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

N a_{2} S i O_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $N a_{4} S i O_{4}$
Number of $N a^{\oplus}$ ions in fcc
$= \frac{1}{2} \times T V = \frac{1}{2} \times 8$ = 4/unit cell
So, formula of solid
$= + 1 \times 4 N a_{4} S i^{+ 4} O_{4}^{2 -} \Rightarrow N a_{4} S i O_{4}$

Suggest Corrections

Similar questions

A series of minerals consists of a crystal in which $F e^{2 +}$ and $M g^{2 +}$ ions may substitute for each other causing a substitutional impurity defect without changing the volume of the unit cell. In the mineral, the oxide ions exist as $f c c$ with $S i^{2 +}$ occupying $\frac{1}{4} t h$ of the octahedral voids and divalent ions occupying $\frac{1}{4} t h$ of the tetrahedral voids. The density of forsterite (Magnesium silicate) is 3.21 $g / c c$ and that of fayalite (Ferrous silicate) is 4.34 $g / c c$ . If the density of mineral is 3.88 g/cc, then which of the following statements is/are CORRECT.

In which of the following structures coordination number for cations and anions in the packed structure will be same ?

(a) $C l^{-}$ ions form fcc lattice and $N a^{+}$ ions occupy all octahedral voids of the unit cell

(b) $C a^{2 +}$ ions form fcc lattice and $F^{-}$ ions occupy all the eight tetrahedral voids of the unit cell

(d) $S^{2 -}$ ions form fcc lattice and $Z n^{2 +}$ ions go into alternate tetrahedral voids of the unit cell

Q. A solid

A^{+} B^{-}

has the

B^{-}

ions arranged in a bcc lattice. If the

A^{+}

ions occupy half of the tetrahedral sites in the structure. The formula of the solid is ?

Q. Oxide ion forming a fcc lattice in which

F e^{2 +}

ions are occupying

1 / 8^{t h}

of tetrahedral void and

F e^{3 +}

ions are occupying

\frac{1}{2}

of the octahedral void. What will be the formula of the compound?

If NaCl is doped with $10^{- 3} m o l %$ of $S r C l_{2^{'}},$ each $S r^{2 +}$ ion replaces two $N a^{+}$ ions, but occupies only one lattice point in place of $N a^{+}$ ion. This creates one cation vacancy.

Join BYJU'S Learning Program

If in forsterite mineral, bivalent Mg2+ ions are to be replaced by unipositive Na⊕ ions, and if Na⊕ ions are occupying half of the TVs in fcc lattice, the arrangement of rest of the constituents is kept same, then the formula of the new solid is:

The correct option is C Na4SiO4Number of Na⊕ ions in fcc=12×TV=12×8 = 4/unit cellSo, formula of solid =+1×4Na4Si+4O2−4⇒Na4SiO4

If in forsterite mineral, bivalent $M g^{2 +}$ ions are to be replaced by unipositive $N a^{\oplus}$ ions, and if $N a^{\oplus}$ ions are occupying half of the TVs in fcc lattice, the arrangement of rest of the constituents is kept same, then the formula of the new solid is:

The correct option is C $N a_{4} S i O_{4}$
Number of $N a^{\oplus}$ ions in fcc
$= \frac{1}{2} \times T V = \frac{1}{2} \times 8$ = 4/unit cell
So, formula of solid
$= + 1 \times 4 N a_{4} S i^{+ 4} O_{4}^{2 -} \Rightarrow N a_{4} S i O_{4}$