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Electronic configuration of O = 1s2 2s2 2p4 and that for N = 1s2 2s2 2p5. Here, in N, the 2p subshell already has a HALF FILLED electronic arrangement unlike in O. This ensures greater stability for N atom than for O atom and thus, ionisation enthalpy (energy required to remove outermost electron) for O becomes lesser in magnitude than that for N.

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The first ionization enthalpy of B is less than first ionization enthalpy of Be . Why?

Posted by Shivani Kumari 3 years, 11 months ago

CBSE > Class 11 > Chemistry

1 answers

Anjan Karthi 3 years, 11 months ago

Be has an Electronic configuration of 1s2 2s2 and B has an electronic configuration of 1s2 2s2 2p1. The p 2p orbital, causing lesser shielding effect than 2s orbital, would be able to loose an electron much more easily than 2s orbital. Hence, Be has greater ionization enthalpy than B.

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Types of chemical reaction in chapter 12

Posted by Dharni Rathor 3 years, 11 months ago

CBSE > Class 11 > Chemistry

1 answers

Shravani Tambe 3 years, 11 months ago

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SIMILARITIES AND DISIMILARITIES BETWEEN HYDROGEN,HALOGEN AND ALKALI METALS

Posted by Apurva J Patil 3 years, 11 months ago

CBSE > Class 11 > Chemistry

1 answers

Prabhjot Kaur 3 years, 11 months ago

Similarities between hydrogen and alkali metals : Electronic configuration Electropositive character Valency and oxidation state Reducing agents Dissimilarities between alkali metals: Ionisation enthlapy Non metallic in nature Atomicity Similarities between hydrogen and halogen Electronic configuration Ionisation enthlapy Electronegative character Oxidation state Diatomic nature Nature of compounds Dissimilarities between hydrogen and halogen Less tendency for hydride formation Absence of unshared pair of electron Oxides of halogens are acidic

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For hydrogen atom the energy is expressed as En=-13.6/n ev/atom. The value of 2nd Ionisation energy of helium atom will be

Posted by Kiran Mandal 3 years, 11 months ago

CBSE > Class 11 > Chemistry

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Calculate the volume of 0.015M HCL solution required to prepare 250ml of a 5.25×1/1000M HCL solution

Posted by Pinki Borah 3 years, 11 months ago

CBSE > Class 11 > Chemistry

2 answers

Kaushik Ka 3 years, 11 months ago

M1*V1 = M2*V2 (0.015×250)÷(5.25÷1,000) = V2 V2 = 714.285 ml

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Kaushik Ka 3 years, 11 months ago

M1*V1 = M2*V2

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Calculate the number of molecules in 20 g aluminum sulphate

Posted by Baazigha Sultana Siddiqui 3 years, 11 months ago

CBSE > Class 11 > Chemistry

1 answers

Anjan Karthi 3 years, 11 months ago

Molar mass of Al2(SO4) 3 = 2(27) + (32 + 64) 3 = 54 + 288 = 342 g. // No of molecules = (20/342) × 6.02 × 10^23 = 0.352 × 10^23 molecules//

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How many moles of methane required to produce 88g of CO2 after combustion

Posted by Akash Malsa 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Anjan Karthi 4 years ago

According to balanced equation, CH4 + 2O2 → CO2 + 2H2O. No of moles of CO2 produced = W2/M2 = 88/44 = 2 mol// Therefore, no. of moles of Methane required = 2 mol. //

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Electronic configuration of florine

Posted by Kanak Sahu 4 years ago

CBSE > Class 11 > Chemistry

3 answers

Khushi Rani 4 years ago

1s*2 2s*2p*5

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Jatin Jatin 4 years ago

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Jatin Jatin 4 years ago

1s² 2s² 2p⁵

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Which out of the following contain least number of atoms? A. 1gm of H2O B. 1gm of CH4 C. 1gm of N2

Posted by Anoubam Victor 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Ningmi Chon 4 years ago

1×4×6•023×1023

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Which out of the following contain least number of atoms?

Posted by Anoubam Victor 4 years ago

CBSE > Class 11 > Chemistry

0 answers

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Calculate the percent ionic character of HCl. Given that the observed dipole moment is 1.03D and length of HCl is 1.275 Å.

Posted by Priyanka Khatri 4 years ago

CBSE > Class 11 > Chemistry

2 answers

Priyanka Khatri 3 years, 11 months ago

Thanks

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Roopam Almadi 4 years ago

Fraction of charge = 1.03 × 10^-18 esu. cm / 4.8 × 10 ^-10 esu × 1.275 × 10 ^-8 cm Fraction of charge = 0.1683 % Ionic character = Fraction of charge × 100 So% ionic character = 16.83 % So % covalent character = 100 - 16.83 = 83.17 % ....

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If 2.36grm of sodium sulphate is added to berium chloride the residue is formed to be 3.88grm of barium sulphate and 1.98grm of sodium chloride.What is the mass of barium chloride?

Posted by Deiba Pariong 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Anjan Karthi 4 years ago

Law of conservation of mass. LHS = RHS. Therefore, 2.36g Na2SO4 + x g BeCl2 = 3.88g BaSO4 + 1.98g NaCl. Therefore, required mass of BeCl2 = 3.5 g. //

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A mixture of mgco3 and caco3 I was heated for a long time the wait is decreased by 50% the percentage composition of mgco3 and that of caco3 in the mixture will be respectively

Posted by Pushpa Kumar 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Himanshu Mali 4 years ago

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Short note on Development of Chemistry

Posted by Baisa Hkm ?? 4 years ago

CBSE > Class 11 > Chemistry

2 answers

Baisa Hkm ?? 4 years ago

What is this

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Amitesh Tiwari 4 years ago

Hdhdjsjvkvvieie

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What is shelding effect?

Posted by Aman Kushwaha 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Anjan Karthi 4 years ago

In case of big atoms like Sulphur and Potassium, the electrons in outer shells is not given a significant part of the effective nuclear charge of the atom due to the inner electrons. This is called SHIELDING EFFECT.

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What are the limitations of octect rule?

Posted by Jissmol Jacob Varghese 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Diksha Thakur 4 years ago

Limitations : (1) Hydrogen with 1 electron attains stability by sharing, gaining or losing 1 valence electron. It does not need to complete octet to attain stability. Also, He has only 2 electrons and is stable. (2) Incomplete octet: In certain molecules such as BeH 2 , BeCl 2 , BH 3 , BF 3 , the central atom has less than 8 electrons in its valence shell, yet the molecule is stable. (3) Expanded octet: In certain molecules such as PF 5 , SF 6 , IF 7 , H 2 SO 4 , the central atom has more than 8 valence electrons, yet the molecule is stable.

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Calculate the number of moles in 7.9mg of calcium

Posted by Devi Devi 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Sameer Kumar Tagore 3 years, 11 months ago

2.0×power of 10 is minus3

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Molecular mass of na2co3

Posted by Khushi Kundal 4 years ago

CBSE > Class 11 > Chemistry

2 answers

Komal Bhanndari 4 years ago

106

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Anjan Karthi 4 years ago

Na2CO3 = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 amu.

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Notes for structure of atom

Posted by Darshan Jain 4 years ago

CBSE > Class 11 > Chemistry

0 answers

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Properties of resonance

Posted by Himanshu Yadav 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Krishna Baranwal 4 years ago

One object can't defined other object

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What weight of ch4 should be taken which contains 2.4 ×10²⁴ atom of H

Posted by Jasmin Singh 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Anjan Karthi 4 years ago

16 g = 4 H × 6.022 × 10^23. X g = 2.4 × 10^24. Therefore, X = 2.4 × 16 × 10^24 / 4 × 6.022 × 10^23 = 15.94 g. //

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5 atomic number

Posted by Gaury Bhardwaj 4 years ago

CBSE > Class 11 > Chemistry

3 answers

Manorma Singh 4 years ago

Boron

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Boss King 4 years ago

.Boron. Z M ⬇ ⬇ 5 10.118u

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Komal Bhanndari 4 years ago

Boron

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Draw ideal gas equation PV=nRT from different gas laws

Posted by 1621 H. Lalrosiami 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Ankit Gupta 4 years ago

Boyle law - V~1/P - (I) equation Charle's Law - V~T. -(ii) equation Avagardo law- V~n. -(iii) equation By usin1,2,3 equation: V~nT/P PV~nT PV=nRT

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What volume of co2 at STP can be obtained on the decomposition of 100 g of caco3

Posted by Jatin Kumar 4 years ago

CBSE > Class 11 > Chemistry

1 answers

Priyanshi Sharma 4 years ago

Volume of co2 at STP is 22.4 L Because molar mass of caco3 = 40 + 12 + 3 × 16 = 100 g. So, co2 = 12 + 2 × 16 = 44 g or 22.4 L

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What is Multiplication and Division of Significant Figures

Posted by Ujjawal Pandey 4 years ago

CBSE > Class 11 > Chemistry

2 answers

Yanshu Kumar 4 years ago

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Shabnam Singh 4 years ago

When multiplying two numbers, the important value is the number of significant figures. ... If the numbers being multiplied have three significant figures, then the product will have three significant figures.

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