12th solve complete question 12 give the formula of each of the following coordination entities: - chemistry - coordination compounds
12th solve complete question

12 give the formula of each of the following coordination
entities:

(i) c o 3 + ion is bound to one c l - ,   o n e   n h 2
and two bidentate ethylene diamine (en) molecules
(ii) n i 2 + ion is bound to two water molecules and two oxalate
ions write the name and magnetic behaviour of each of the above
coordination entities (atomic no of co = 27, ni=28) foreign
2012
dear student

kindly refer to the link for the similar query :


https://www meritnation
com/ask-answer/question/q-23-and-24-q23-give-the-formula-of-each-of-the-follow/the-p-block-elements/11940305

regards
a) he helium (he) is an element by itself. he atom can only exhibit dispersion forces.b) hclh          1a                   1 × 1 e– = 1 e–cl         7a                   1 × 7 e– = 7 e–   total: 8 valence e–[readmore]cl is more electronegative than h, so hcl is a polar covalent compound. hcl exhibits dipole-dipole interaction.c) h2oh          1a                   1 × 2 e– = 2 e–o          6a                   1 × 6 e– = 6 e–   total: 8 valence e– since hydrogen is bonded directly to oxygen.h2o exhibits hydrogen bonding.d) cl2cl         7a                   1 × 7 e– = 14 valence e–composed of the same atom. there's no electronegativity difference.cl2 can only exhibit dispersion forces. we are going to do the following steps:step 1: determine the classification of the compoundstep 2: write the formula of the compound[readmore]step 1: determine the classification of the compoundrecall the definition of the following:ionic compound - usually made up of a metal and a non-metalcovalent compound - composed of more than one nonmetal togetheracid - substance whose molecules yield hydrogen ions (h+) when dissolved in waterhydrate - substance that contains water or its constituent elementsfor cobaltic iodidecobaltic – metaliodide – nonmetalmetal + nonmetal = ionicstep 2: write the formula of the compoundthe rules for naming ionic compounds are as follows:1. the metal keeps its name and is named and written first.– for transition metals, we use roman numerals to denote its charge.2. the nonmetal keeps its base name but has its ending changed to –ide.for cobaltic iodidecobaltic  → transition metal → +2 charge → co2+iodide → group 7a → -1 charge → i-criss-cross the charges and make them the subscript of the other ioncobaltic iodide → coi2answer: cobaltic iodide is an ionic compound with a formula of coi2 a) van der waals (present, all molecule/atoms contain it)[readmore](b) dipole-dipole (present, c=o creates a partial charge on c and o)(c) hydrogen bonding (not present, h-bonds are only present in compounds with h attached on o, n or f)(d) ion-ion (not present, only appears on compound with full charges) we’re being asked to determine the conjugate acid of o2-.recall that the conjugate acid of a base is simply the compound formed when a base gains a proton (h+). it's considered as an acid because it can lose h+ to reform the base.[readmore]for o2–, simply add a hydronium ion from the formula. note that we also need to add the +1 charge of the proton (h+) to o2–, giving our conjugate acid a net charge = -1.o2– + h+ → oh–  –2    +   1     =       –1this means the conjugate acid of o2– is oh–. we are being asked to classify the following reactions based on the following categories:combination/synthesis → multiple reactants combine to form a single productw/o change in oxidation numbersa + b → abdecomposition →reactant splits into multiple products with no change in oxidation numbersab → a + bcombustion reaction: reaction with oxygen that produces carbon dioxide and waterhydrocarbon + o2 → co2 + h2osingle displacement →single element displaces another element in a compoundab + c → ac + b[readmore]let’s analyze the given reactionsa. fe2(co3)2 (s) → fe2o3(s) + 2co2(g) break down of fe2(co3)2(s) to its component ions → decomposition reactionanswer: this is a decomposition reaction.b. 2 c6h11oh(l) + 17 o2(g) → 12co2 (g) + 12h2o(l) reaction with oxygen that produces carbon dioxide and water → combustion reactionanswer: this is a combustion reaction.c. p4o10(s) + 6h2o(l) → 4h3po4(aq) two or more substances combine to form a single compound → combination reactionanswer: this is a combination reaction.d. 2c(s) + mno2(s) → mn(s) + 2co(g) substitute one element for another element in a compound →single-displacement reaction answer: this is a single-displacement reactione. 2naclo3 (s) → 2nacl(s) + 3o2(g)break down of compound into its simpler substances → decomposition reactionanswer: this is a decomposition reaction a) 3ag + 4hn03 > 3agno3 + no + 2h2o

b) mno2 + 4hcl > 2h2o + mncl2 + cl2 <p>concepts and reason</p>
<p>the concept used to solve this problem is the formal charge calculation of the oxygen atoms present in the ozone molecule.</p>
<p>fundamentals</p>
<p>a charge that is assigned to an atom, which is present in the molecule, is represented as a formal charge. in two bonded atoms, a formal charge assumes that the electrons are equally shared between them. the formal charge is calculated using the following formula:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>here, \(\mathrm{v}\) is the number of valence electrons, \(\mathrm{n}\) is the number of non-bonding electrons and \(\mathrm{b}\) is the number of bonding electrons.</p>
<p> </p>
<p>the structure is as follows:</p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p><img src="https://img.wizedu.com/images/5af6d1f54822b4dfb8fe89d91ef2011c5fe3020815aa4.png" alt="" width="88" /></p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(\begin{aligned}=& 6-6-\frac{2}{2} \\ &=-1 \end{aligned}\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the marked oxygen atom has 6 valence electrons. it contains 6 non-bonding electrons. the marked oxygen atom is bonded with one oxygen atom. therefore, the bonding electron of oxygen is \(2 .\) substitute the values in a formal charge formula to determine the formal charge of oxygen (marked).</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/39493056a8515708b79350d1976e17b35fe30216f0473.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-2-\frac{6}{2}\)</p>
<p>\(=+1\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the oxygen atom in the middle has 6 valence electrons and has 2 non-bonding electrons. the central oxygen atom is bonded with 1 double bond and 1 single bond. so, there is a total of 3 bonds containing 6 electrons (bonding electrons). the formal charge of the middle oxygen atom can be determined by substituting the values in a formal charge formula.</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/b1122ce7723aa4ab1f9ae5c9e457299e5fe3022ca77a5.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-4-\frac{4}{2}\)</p>
<p>\(=0\)</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p>
<p>the oxygen atom present on the right side contains 6 valence electrons which have 2 non-bonded electrons. the marked oxygen is bonded with 1 double bond with a central oxygen atom. so, the number of bonding electrons is \(4 .\) the formal charge of the corner oxygen atom (right corner) can be determined by substituting the values in the formal charge formula.</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p> according to ncert answer would be option (1)and

but according to general chemistry answer would be (3)andoption 1)and





option 2)andoption 3)andoption 4)and cao we are asked to write the chemical formula of hydrates and to use proper symbols, case, and formatting.remember that an ionic hydrate is formed when an ionic compound is bonded to one or more molecules of water. naming ionic hydrates has the same set of rules in naming the ionic compound portion. the water molecule portion is named with a numerical prefix together with the word hydrate.[readmore]step 1: discuss the rules in naming ionic hydrates.step 2: from the naming rules, extract the chemical symbols and the formula for each of the compounds.step 1: discuss the rules in naming ionic hydrates.rules in naming the ionic compound portion:1.a. the metal is named and normally written first.-if the metal is a transition metal we must use a roman numeral to describe its positive charge.1.b. the nonmetal keeps the first part of its name but has its ending changed to -ide.1.c. name the polyatomic as you would normally.rules in naming the h2o portion:1.d. the h2o portion will be called hydrate.1.e. to describe the number of h2o molecules, use the numerical prefixes.note: a dot (•) is written in between the ionic compound and the hydrate.step 2: from the naming rules, extract the chemical symbols and the formula for each of the compounds.a. iridium(iii) bromide tetrahydrate rule 1.a.iridium→transition element → symbol : iriii → describes the transition element's positive charge → symbol: 3+rule 1.b.bromide → nonmetal bromine → bromide ion has a charge of -1 → symbol: brwe write the metal and nonmetal beside each other. recall that the charges will be exchanged, becoming subscripts in the chemical formula.we don't need to write 1 as a subscript, it is understood to be 1 without it.rule 1.d.hydrate → h2orule 1.e.tetra → four → 4therefore the chemical formula is irbr3•4h2o.b. tin(iv) chloride pentahydraterule 1.a.tin(iv)→ sn4+rule 1.b.chloride → cl-we write the metal and nonmetal beside each other. recall that the charges will be exchanged, becoming subscripts in the chemical formula.again, we don't need to write 1 as a subscript.rules 1.d and 1.e.pentahydrate → 5 h2o the chemical formula is sncl4•5h2o.therefore the chemical formula is sncl4•5h2o.c. zinc nitrate hexahydrate rule 1.a.zinc→ zn1+rule 1.c. nitrate → no31-we write the metal and nonmetal beside each other. recall that the charges will be exchanged, becoming subscripts in the chemical formula.note that +1 and -1 charges just cancel out, we just put the atoms together.rules 1.d and 1.e.hexahydrate → 6 h2othe chemical formula is znno3•6h2o.therefore the chemical formula is znno3•6h2o.d. sodium carbonate monohydraterule 1.a.sodium→ na1+rule 1.c. carbonate → co32-we write the metal and nonmetal beside each other. recall that the charges will be exchanged, becoming subscripts in the chemical formula.again, we don't need to write 1 as a subscript.rules 1.d and 1.e.monohydrate → 1 h2o → h2o     (we don't need to write 1 as a coefficient.)the chemical formula is na2co3•h2o.therefore the chemical formula is na2co3•h2o. i. given:valency of carbon is 4valency of chlorine is 1by applying criss-cross method:formula: ccl4thus, the formula for carbon tetrachloride is ccl4.ii. given:valency of element x is 1valency of oxygen is 2by applying criss-cross method:formula: x2o