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Question

Match the amount of reactant given in column I with neutralisation reactions given in column II.

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Solution

(A) 1. mEq of NaOH =200×0.5=100mEq100mEq of H2SO4
2. 4.9 g H2+6SO4=4.998×103
=50 mmol100mEq (n factor =2)
200 mmol of O atoms. (1molH2SO44molOatoms)
3.. Highest oxidation state of S=+6
4. Hybridisation=sp3; geometry and shape Tetrahedral
(B) 4.9 g H3PO44.998×1000=50 mmol
=50×3 (n factor)150 mEq
2.50 mmol of H3PO450×4=200 mmol of O atoms).
(1 mol H3PO44mol O atoms)
1. But 150 mEq H3PO4mEq of NaOH
5. Hybridisation is sp3, so geometry and shape is tetrahedral.
3. Maximum oxidation state of P=+5
(C) 1.4.9 g H2C2O44.990×1000
=50 mmol=100 mEq (n factor =2)
2. 50 mmoles of H2C2O450×4=200 mmolesO atoms
(1 mol H2C2O4=4mol o atoms)
100mEqNaOH
3. Highest oxidation state of C=+4
4. Reacts with oxidising agents such as MNO4 and Cr2O27.
(D) 1.5.3 g of Na2CO)3=5.3106×1000
=50 m moles
2 Na2CO3 does not react with NaOH.
3. Highest oxidation states of C=+4
5.CO23, hybridisation is sp2, so geometry and shape is planar.
7. 1 mol Na2CO33mol O atoms.

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