Neet formula 2

 NEET JEE CBSE BIOLOGY CHEMISTRY:

Trigonometric Ratio Trick -

Pandits Badri Prasad, Har Har Bole, Sona Chandi Tole

P/H = Sin, B/H = Cos, P/B = Tan

Trigonometric Signs in Quadrants  Trick -

All Silver Tea Cups

Quadrant 1 - All Positive

Quadrant 2 - sin positive, therefore cosec

Quadrant 3 - tan positive, therefore cot

Quadrant 4 - cos positive, therefore sec

🔰INORGANIC IN SHOTS🔰

✌️Increasings or Decreasing Order

✨01. Melting point=

Li > Na > K > Rb > Cs

✨02. Colour of the flame=

Li-Red, Na-Golden, K-Violet, Rb-Red, Cs-Blue, Ca-Brick red, Sr-Blood red, Ba-Apple green

✨03. Stability of hydrides =

LiH > NaH > KH > RbH> CsH

✨04. Basic nature of hydroxides=

LIOH < NaOH < KOH < RbOH < CsOH

✨05. Hydration energy=

Li> Na > K> Rb > Cs

✨06.) Reducing character=

Li > Cs > Rb > K > Na

✨07. Stability of +3 oxidation state=

 B> Al > Ga > In > T1

✨08. Stability of +1 oxidation state= Ga < In < TI

✨09. Basic nature of the oxides and hydroxides= 

B< Al< Ga < In < TI

✨10. Relative strength of Lewis acid= BF3 < BCl3 < BBr3 < BI3

✨11. Ionisation energy=

B> Al  In  SiO2 > Ge02 > SnO2 > PbO2

✨15. Reducing nature of hydrides=

 CH4 < SiH4  < GeH4  < SnH4 < PbH4

✨16. Thermal stability of tetrahalides=

CCl4> SiCl4> GeCl4> SnCl4 > PbCl4

✨17. Oxidising character of M+4 species=

 GeCl4 < SnCl4 < PbCl4

✨18. Ease of hydrolysis of tetrahalides=

 SiCl4 < GeCl4 < SnCl4 < PbCI4

✨19. Acidic strength of trioxides=

 N203 > P2O3 > As2O3

✨20. Acidic strength of pentoxides=

N2O2 > P2O2> As202 > Sb2O2 > Bi‌202

✨21) Acidic strength of oxides of nitrogen= 

N2O < NO  PH3 > AsH3 > SbH3 > BiH3

✨23. Stability of trihalides of nitrogen=

 NF3 > NCl3 > NBr3

✨24.Lewis base strength=

NF3  PCI3 > AsCl3 > SbCl3 > BiCl3

✨26.Lewis acid strength of trihalides of P, As, and Sb=

PCl3 > ASCl3 > SbCl3

✨27. Lewis acid strength among phosphorus trihalides

PF3 > PCl3 > PBr3 > PI3

✨(28) Melting and boiling point of hydrides=

 H2O > H2Te > H2Se >H2S

✨29. Volatility of hydrides=

H2O < H2Te < H2Se < H2S

✨30. Reducing nature of hydrides=

H2S < H2Se < H2Te

✨31. Covalent character of hydrides=

 H2O < H2S < H2Se < H2Te

✨32. The acidic character of oxides (elements in the same oxidation state)=

SO2 > SeO2 > TeO2 > PoO2

 SO3 > SeO3  > TeO3

✨33. Acidic character of oxide of a particular element (e.g. S)=

SO < SO2 < SO3

SO2 > TeO2 > SeO2 > PoO2

✨34. Bond energy of halogens=

Cl2 > Br2 > F2 > I2

✨35. Solubility of halogen in water =

F2 > Cl2 > Br2 > I2

✨36. Oxidising power=

F2 > Cl2 > Br2 > I2 

✨37. Enthalpy of hydration of X ion=

F- > Cl- > Br- >I-

✨38. Reactivity of halogens:=

F> Cl> Br > I

✨39. Ionic character of M-X bond in halides

= M-F > M-Cl > MBr > M-I

✨40. Reducing character of X ion:=

I- > Br- > Cl- > F-

✨41. Acidic strength of halogen acids=

HI > HBr > HCI > HF

✨42 Reducing property of hydrogen halides

= HF < HCL < HBr < HI

✨43. Oxidising power of oxides of chlorine

= Cl2O > ClO2 > Cl206 > Cl2O7

✨44. Decreasing ionic size= 

02- > F- > Na+ > Mg2+

✨45 Increasing acidic property=

Na2O3 < MgO < ZnO< P205

✨46 Increasing bond length=

N2 <02 < F2 < CL2

🌺Revision Notes on Sexual Reproduction in Flowering Plants🌺

(3) Entry of pollen tube into embryo sac:

The pollen tube enters the embryo sac only from the micropylar end irrespective of its mode of entry into the ovule. The pollen tube either passes between a synergid and the egg cell or enters into one of the synergids through filiform apparatus. The synergids direct the growth of pollen tube by secreting some chemical substances (chemotropic secretion). The tip of pollen tube enters into one synergid. The penetrated synergid starts degenerating. After penetration, the tip of pollen tube enlarges and ruptures releasing most of its contents including the two male gametes and the vegetative nucleus into the synergid.

(4) Double fertilization:

The nuclei of both the male gametes are released in the embryo sac. One male gamete fuses with the egg to form the diploid zygote. The process is called syngamy or generative fertilization. This syngamy was discovered by Strasburger (1884). The diploid zygote finally develops into embryo. The other male gamete fuses with the two polar nuclei (or secondary nucleus) to form the triploid primary endosperm nucleus. The process is called triple fusion or vegetative fertilization. These two acts of fertilizations constitute the process of double fertilization. The process was discovered by S.G. Nawaschin (1898) and Guignard in Lilium and Frittillaria. Double fertilization occurs in angiosperms only.

✨47 Increasing size=

Ca2+ < Cl- < S2-

✨48 Increasing acid strength=

 HClO < HClO2 < HClO3 < HClO4