Inorganic is'nt always Memorizing

Chemistry Level 4

Say we have the following compound

( S i X 7 O X 17 ) ( O H ) X 4 K X 2 F e X I I X x F e X I I I X y \ce{(Si7O17)(OH)4K2Fe^{II}_{x}Fe^{III}_{y}}

Now take the value of x × y x \times y .

Which of these is/are equal to x × y x \times y

  1. The number of double bonds present in A l X 4 C X 3 \ce{Al4C3}

  2. The square root of the number of 2 c 2 e 2c - 2e bonds present in B X 2 H X 6 \ce{B2H6}

  3. The number of B O B B-O-B present in N a X 2 B X 4 O X 7 \ce{Na2B4O7}

  4. The square root of the number of B O H \ce{B-OH} present in N a X 2 B X 4 O X 7 \ce{Na2B4O7} complex.

  5. The cube root of the number of water of crystallistation present in N a X 2 B X 4 O X 7 10 H X 2 O \ce{Na2B4O7.10H2O} .

  6. The number of triple bonds present in A l X 4 C X 3 \ce{Al4C3}

Enter the answer as a product of the option numbers.

Clarification

  • If answers are 1,2,3 then enter the value 1 × 2 × 3 = 6 1 \times 2 \times 3 = \boxed{6}

  • Here F e X I I and F e X I I I \ce{Fe^{II}} \text{and} \space \ce{Fe^{III}} are Ferrous(II) ion and Ferric(III) ion.


This question made by me is Purely Original from the chapter p block elements



Try my set Block Chemistry


The answer is 240.

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1 solution

Md Zuhair
Jan 7, 2018

Relevant wiki: Inorganic chemistry

Lets try this!

Okay...

So We know that the general formula for cyclic/ring silicates is ( S i X n O X 3 n ) X 2 n \ce{(Si_{n}O_{3n})^{-2n}}

Now lets make the term a bit more confined...

( S i X 7 O X 17 ) ( O H ) X 4 K X 2 F e X I I X x F e X I I I X y ( S i X 7 O X 21 ) H X 4 K X 2 F e X I I X x F e X I I I X y \ce{(Si7O17)(OH)4K2Fe^{II}_{x}Fe^{III}_{y}}\ \implies \ce{(Si7O21)H4K2Fe^{II}_{x}Fe^{III}_{y}}

So for cyclic silicates with n=7 we get the charge for S i X 7 O X 21 \ce{Si7O21} which is nothing but 14 -14 .

Now we have oxidation for number of H andK \ce{H} \space \space \text{and} \text{K} to be + 1. +1.

Again for F e X I I X x and F e X I I I X y \ce{Fe^{II}_{x}} \text{and} \ce{Fe^{III}_{y}} are +2 and +3 resp.

So we have the oxidation number of total compound = 0 =0

2 x + 3 y 14 + 4 + 2 = 0 \implies 2x+3y-14+4+2=0

2 x + 3 y = 8 \implies 2x+3y=8

y = 2 and x = 1 \implies y=2 \space \text{and} \space x=1 ...

So we get x × y = 2 x \times y =2

Now the options are structure related.

So

  1. There are 2 double bonds in A l X 4 C X 3 \ce{Al4C3} [ C O R R E C T ] [CORRECT]

  2. There are 4 2c-2e bonds in B X 2 H X 6 \ce{B2H6} [ C O R R E C T ] [CORRECT]

  3. There are 5 B O B \ce{B-O-B} bonds present. [ W R O N G ] [WRONG]

  4. There are 4 B-OH bonds present in N a X 2 B X 4 O X 7 \ce{Na2B4O7} complex. [ C O R R E C T ] [CORRECT]

  5. The water of crystallisation in N a X 2 B X 4 O X 7 \ce{Na2B4O7} is 8. [ C O R R E C T ] [CORRECT]

  6. Number of triple bonds in A l X 4 C X 3 \ce{Al4C3} is 2. [ C O R R E C T ] [CORREC T]

Hit and trail worked there, if I assume another point on that line for which x y x * y gives another value then how will you solve it?

Sahil Silare - 3 years, 2 months ago

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i didnt got your query. can you please say it clearly?

Md Zuhair - 3 years, 2 months ago

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If I took x = 4 x=4 and y = 0 y=0 then the answer would be none of these.

Sahil Silare - 3 years, 2 months ago

Isnt Al4C3 an ionic compound? how can it have double/triple bonds?

Adarsh Agrawal - 3 years, 2 months ago

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No no. Its an covalent compound. You can see the structure in net!

Md Zuhair - 3 years, 2 months ago

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