Make placards with symbols and valencies of the atoms of the element separately.Each student should hold two placards, one with the symbol in the right hand and the other will the valency in the left hand. Keeping the symbols in place , students should criss-cross their valencies to form the formula of a compound.

The equivalent weight of a species if acts as oxidant or reductant should be derived by: Eq. weight Mol.wt. of oxidant or reductant = ("Mol.w.t of oxidant or reductant")/("Number of electrons lost or gained by onemclecule of oxidant or reductant") . During chemical reactions, equal all equivalents of one species react with same number of equivalents of other species giving same number of equivalent of products. However this is not true for reactants if they react in terms of moles. Also Molarity can be converted to normality by multiplying the molarity with valence factor or .n. factor The equivalent weight of an element is 13.16. It forms an acidic oxide which with KOH forms a salt isomorphous with K_(2)SO_4 The atomic weight of element is :

Let us do the following activity to understand the mendelian principles of heredity. Materaials required: a. 3cm length and 1 cm breadth chart pieces -4 b. 2 cm length and 1 cm breadth chart pieces -4 c. Red byuttons -4 d. white buttons -4 e. chart, scale, sketch pen penil, 2 bags. Method: Prepare a chart with 2x2 boxes along with numbe and symbol as shown in the figure. Game 1: Monohybrid cross (starting with hybrid parents) To start with take 1,2 or 3,4 . In case you start 1,2 pik all the 16 log and short pieces and prepare such paris in each of which you have a long and short piece. Take 4 pairs each of long and short strips and put them in two separate bags. Now each bag contains 8 strips (4 long and 4short). One bag say 'A' represents mael and the bag 'B' represents female . Now rendomly pick one strip each from bag A and B and put them together in the 1 on the chart. Keep picking out the strips and arrange them in the same manner till your bags are empty. Same time your boxes in the chart are filled with paris of strips. you might have got the following combinations, two long strips, one long and one short strip, two short strips. (Q) B. What is the number of one long and one short pairs?

Let us do the following activity to understand the mendelian principles of heredity. Materaials required: a. 3cm length and 1 cm breadth chart pieces -4 b. 2 cm length and 1 cm breadth chart pieces -4 c. Red byuttons -4 d. white buttons -4 e. chart, scale, sketch pen penil, 2 bags. Method: Prepare a chart with 2x2 boxes along with numbe and symbol as shown in the figure. Game 1: Monohybrid cross (starting with hybrid parents) To start with take 1,2 or 3,4 . In case you start 1,2 pik all the 16 log and short pieces and prepare such paris in each of which you have a long and short piece. Take 4 pairs each of long and short strips and put them in two separate bags. Now each bag contains 8 strips (4 long and 4short). One bag say 'A' represents mael and the bag 'B' represents female . Now rendomly pick one strip each from bag A and B and put them together in the 1 on the chart. Keep picking out the strips and arrange them in the same manner till your bags are empty. Same time your boxes in the chart are filled with paris of strips. you might have got the following combinations, two long strips, one long and one short strip, two short strips. (Q) D. What is the percentage of each type? also find their ratios.

Let us do the following activity to understand the mendelian principles of heredity. Materaials required: a. 3cm length and 1 cm breadth chart pieces -4 b. 2 cm length and 1 cm breadth chart pieces -4 c. Red byuttons -4 d. white buttons -4 e. chart, scale, sketch pen penil, 2 bags. Method: Prepare a chart with 2x2 boxes along with numbe and symbol as shown in the figure. Game 1: Monohybrid cross (starting with hybrid parents) To start with take 1,2 or 3,4 . In case you start 1,2 pik all the 16 log and short pieces and prepare such paris in each of which you have a long and short piece. Take 4 pairs each of long and short strips and put them in two separate bags. Now each bag contains 8 strips (4 long and 4short). One bag say 'A' represents mael and the bag 'B' represents female . Now rendomly pick one strip each from bag A and B and put them together in the 1 on the chart. Keep picking out the strips and arrange them in the same manner till your bags are empty. Same time your boxes in the chart are filled with paris of strips. you might have got the following combinations, two long strips, one long and one short strip, two short strips. (Q) A. what is the number of long strip paris?

Let us do the following activity to understand the mendelian principles of heredity. Materaials required: a. 3cm length and 1 cm breadth chart pieces -4 b. 2 cm length and 1 cm breadth chart pieces -4 c. Red byuttons -4 d. white buttons -4 e. chart, scale, sketch pen penil, 2 bags. Method: Prepare a chart with 2x2 boxes along with numbe and symbol as shown in the figure. Game 1: Monohybrid cross (starting with hybrid parents) To start with take 1,2 or 3,4 . In case you start 1,2 pik all the 16 log and short pieces and prepare such paris in each of which you have a long and short piece. Take 4 pairs each of long and short strips and put them in two separate bags. Now each bag contains 8 strips (4 long and 4short). One bag say 'A' represents mael and the bag 'B' represents female . Now rendomly pick one strip each from bag A and B and put them together in the 1 on the chart. Keep picking out the strips and arrange them in the same manner till your bags are empty. Same time your boxes in the chart are filled with paris of strips. you might have got the following combinations, two long strips, one long and one short strip, two short strips. (Q) C. What is the number of short strip pairs?

Let us do the following activity to understand the mendelian principles of heredity. Materaials required: a. 3cm length and 1 cm breadth chart pieces -4 b. 2 cm length and 1 cm breadth chart pieces -4 c. Red byuttons -4 d. white buttons -4 e. chart, scale, sketch pen penil, 2 bags. Method: Prepare a chart with 2x2 boxes along with numbe and symbol as shown in the figure. Game 1: Monohybrid cross (starting with hybrid parents) To start with take 1,2 or 3,4 . In case you start 1,2 pik all the 16 log and short pieces and prepare such paris in each of which you have a long and short piece. Take 4 pairs each of long and short strips and put them in two separate bags. Now each bag contains 8 strips (4 long and 4short). One bag say 'A' represents mael and the bag 'B' represents female . Now rendomly pick one strip each from bag A and B and put them together in the 1 on the chart. Keep picking out the strips and arrange them in the same manner till your bags are empty. Same time your boxes in the chart are filled with paris of strips. you might have got the following combinations, two long strips, one long and one short strip, two short strips. (Q) E. What can you coclude from this game?

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition must compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) the permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The orbital angular momentum of an electron in p-orbital makes an angle of 45^@ from Z-axis. Hence Z-component of orbital angular momentum of election is :

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition must compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) he permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The maximum orbital angular momentum of an electron with n= 5 is

It is tempting to think that all possible transitions are permissible, and that an atomic spectrum arises from the transition of the electron from any initial orbital to any other orbital. However, this is not so, because a photon has an intrinsic spin angular momentum of sqrt2 (h)/(2pi) corresponding to S = 1 although it has no charge and no rest mass. On the other hand, an electron has got two types of angular momentum : Orbital angular momentum, L=sqrt(l(l+1))h/(2pi) and spin angular momentum, arising from orbital motion and spin motion of electron respectively. The change in angular momentum of the electron during any electronic transition mush compensate for the angular momentum carries away by the photon. to satisfy this condition the difference between the azimuthal quantum numbers of the orbital within which transition takes place must differ by one. Thus, an electron in a d-orbital (1 = 2) cannot make a transition into an s = orbital (I = 0) because the photon cannot carry away enough angular momentum. An electron as is well known, possess four quantum numbers n, I, m and s. Out of these four I determines the magnitude of orbital angular momentum (mentioned above) while (2n m determines its z-components as m((h)/(2pi)) the permissible values of only integers right from -1 to + l. While those for I are also integers starting from 0 to (n − 1). The values of I denotes the sub shell. For I = 0, 1, 2, 3, 4,..... the sub-shells are denoted by the symbols s, p, d, f, g, .... respectively The spin-only magnetic moment of free ion is sqrt(8) B.M. The spin angular momentum of electron will be

An orbital is designated by certain values of first three quantum numbers (n, l and m) and according to Pauli.s exclusion principle, no two electrons in a atom can have all the for quantum numbers equal. N, l and m denote size, shape and orientation of the orbital. The permissible values of n are 1,2,3.... prop while that of 1 are all possible integral values from 0 to n-n. Orbitals with same values of n and 1 but different values of m (where m can have any integral values from 1 to +1 including zero) are of equal energy and are called degenerate orbitals. However degeneracy is destroyed in homogeneous external magnetic field due to different extent of interaction between the applied field and internal electronic magnet of different orbitals differing in orientations. In octahedral magnetic field external magnetic field as oriented along axes while in tetrahedral field the applied field actas more in between the axes than that on the axes themselves. For 1=0, 1,2,3,...., the states (called sub-shells) are denoted by the symbol s,p,d,f.....respectively. After f, the subshells are denoted by letters alphabetically 1 determines orbital angular motion (L) of electron as L = sqrt(l(l+1))(h)/(2pi) ON the other hand, m determines Z-component of orbital angular momentum as L_(Z) = m((h)/(2pi)) Hund.s rule states that in degenerate orbitals electrons do not pair up unless and until each each orbitals has got an electron with parallesl spins Besides orbital motion,an electron also posses spin-motion. Spin may be clockwise and anticloskwise. Both these spin motions are called two spins states of electrons characterized by spin Q.N (s) : s = +(1)/(2) and = -(1)/(2) respectively The sum of spin Q.N. of all the electrons is called total spin(s) and 2s+1 is called spin multiplicity of the configuration as a whole. The spin angular momentum of an electron is written as L_(s) = sqrt(s(s+1))(h)/(2pi) The orbital angular momentum of electron (l=1) makes an angles of 45^(@) from Z-axis. The L_(z) of electron will be